COSMOSM Basic FEA System

About the Verification Problems...

. . . . . . . . . . . . . . . . . 1-1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1

Linear Static Analysis

. . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

Contents

COSMOSM Basic FEA System

COSMOSM Basic FEA System

Part 2 Verification Problems

Contents

COSMOSM Basic FEA System

Modal (Frequency) Analysis

. . . . . . . . . . . . . . . . . . . . . . 3-1

COSMOSM Basic FEA System

Part 2 Verification Problems

Buckling Analysis

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4-1

Contents

COSMOSM Basic FEA System

Index

COSMOSM Basic FEA System

1-1

About the Verification
Problems...

Introduction

COSMOSM software modules are continually in the process of extensive develop-
ment, testing, and quality assurance checks. New features and capabilities incorpo-
rated into the system are rigorously tested using verification examples and in-house
quality assurance problems. All verification problems are provided to the user
along with the software, and they are made available in the COSMOSM directory.
There are more than 150 verification problems for analysis modules in the Basic
System.

The purpose of this section is dual fold: to present many example problems that test
a combination of capabilities offered in the COSMOSM Basic System, and to pro-
vide a large number of verification problems that validate the basic modeling and
analysis features. The first part of this manual presented several fully described and
illustrated examples which cover few aspects of modeling and analysis limitations.
This part provides examples on many other analysis features of the Basic System.

The input files for all verification problems are provided in separate folders
(depending on the analysis type) in the “...\Vprobs” directory where “...” denotes
the COSMOSM directory.

Chapter 1 About the Verification Problems...

1-2

COSMOSM Basic FEA System

Folder

Analysis Type

Geostar

GEOSTAR modeling examples

Buckling

Linearized buckling analysis

AdvDynamics Linear dynamic response analysis

Emagnetic

Electromagnetic analysis

Frequency

Frequency (modal) analysis

Fatigue

Fatigue analysis

Nonlinear

Nonlinear dynamic analysis

Static

Linear static stress analysis

Thermal

Thermal (heat transfer) analysis (linear)

FFE

FFE modules

HFS

High frequency electromagnetic simulation

To use the verification problems, enter GEOSTAR and at the GEO> prompt, exe-
cute the command

Load... (FILE)

from the File menu. The following pages show a

listing of the verification problems based on analysis and element type’s.

Classification by Analysis Type

Analysis

Type

Folder

Problem Title

Linear
Static
Analysis

...\Vprobs\Static

S1, S2, S3A, S3B, S4, S5, S6, S7, S8, S9A, S9B,S10A,
S10B, S11, S12, S13, S14A,S14B, S15, S16A, S16B,
S17, S18, S19, S20, S21A, S21B, S22, S23, S24, S25,
S26, S27, S28, S29, S29A, S30, S31, S31A, S32A, S32B,
S32C, S32D, S32M, S33A, S33M, S34, S35A, S35B,
S36A, S36M, S37, S38, S39, S40, S41, S42, S43, S44A,
S44B, S45, S46, S46A, S46B, S47, S47A, S47B, S48,
S49A, S49B, S50A, S50B, S50C, S50D, S50F, S50G,
S50H, S51, S52, S53, S54, S55, S56, S57, S58, S58B,
S59A, S59B, S59C, S60, S61, S62, S63, S64A, S64B,
S65, S66, S67, S68, S69, S70, S71, S74, S75, S76, S77,
S78, S79, S80, S81, S82, S83, S84, S85, S86, S87,

Buckling
Analysis

...\Vprobs\Buckling

B1, B2, B3, B4, B5A, B5B, B6, B7A, B7B, B8, B9, B10,
B11, B12, B13, B14, B15A, B15B

Modal
Analysis

...\Vprobs\Frequency

F1, F2, F3, F4, F5, F6, F7, F8, F9, F10, F11A, F11B, F12,
F13, F14, F16A, F16B, F17, F18, F19, F20A, F20B, F20C,
F20D, F20, F20E, F20G, F20F, F21, F22, F23, F24, F25,
F26, F27, F28

COSMOSM Basic FEA System

1-3

Part 2 Verification Problems

Classification by Element Type

Element

Name

Analysis Type

Problem Title

BEAM2D

Buckling
Linear Static

B10, B11, B13
S9A, S9B, S24, S41, S46, S47, S51, S52, S53, S54, S75,
S76

BEAM3D

Buckling
Modal Analysis
Linear Static

B1, B2, B3, B6, B9
F3, F4, F5, F12, F17
S7, S22, S23, S26, S27, S28, S33A, S33M, S34, S39, S43,
S45, S55

BOUND

Linear Static

NONE

ELBOW

Linear Static

S15, S16A, S16B

GAP

Linear Static

S75, S76

GENSTIF

All

NONE

MASS

Modal Analysis
Linear Static

F1, F5, F6
S39

PIPE

Modal Analysis
Linear Static

F6
S16A, S16B

PLANE2D

Modal Analysis
Linear Static

F2, F20A, F20B, F21, F23
S2, S5, S6, S17, S19, S38, S46, S46A, S48, S49A, S50A,
S50B, S50C, S61, S62, S65, S66, S67, S68, S70, S76,
S82, S83, S86

RBAR

Linear Static

F28

SHELL3

Buckling
Modal Analysis
Linear Static

B5
F11
S3A, S3B, S8, S30, S33A

SHELL3L

Linear Static

NONE

SHELL3T

Modal Analysis

F8, F16A

SHELL4

Buckling
Modal Analysis
Linear Static

B4, B7, B9
F7, F9, F10, F18, F27A, F27B
S20, S25, S33A, S33M, S36A, S36M, S42, S44A, S44B,
S85

SHELL4L

Linear Static

S21A, S31, S43, S59B, S71

Chapter 1 About the Verification Problems...

1-4

COSMOSM Basic FEA System

Classification by Element Type (Concluded)

)

Element

Name

Analysis Type

Problem Title

SHELL4T

Modal Analysis
Linear Static

F16B
S50C

SHELL9

Linear Static

S46B, S56, S57, S58, S59A, S60

SHELL9L

Linear Static

S21B, S29A, S31A, S59A

SHELLAX

Buckling
Modal Analysis
Linear Static

B8, B14, B15
F19, F25, F26
S18, S37, S79, S80, S81

SOLID

Modal Analysis
Linear Static

F13, F20E, F20F, F22
S10A, S10B, S11, S35A, S47, S47B, S49B, S50F, S50G,
S77

SHELL6

Buckling
Modal Analysis
Linear Static

B5B, B7B
F7B, F11B, F20H
S6B, S20B, S42B, S50I

SOLIDL

Linear Static

S29, S35B, S59C

SOLIDPZ

Modal Analysis

F24

SPRING

Modal Analysis

NONE

TETRA10

Modal Analysis
Linear Static

F20D
S50E

TETRA4

Linear Static

NONE

TETRA4R

Linear Static
Modal Analysis

S50H, S58B, S74
F20G

TRIANG

Modal Analysis
Linear Static

F20C
S50D, S64A, S64B, S68, S69, S78, S84

TRUSS2D

Buckling
Modal Analysis
Linear Static

B6
F14
S4, S32A, S32B, S32C, S32D, S32M, S40, S63, S76

TRUSS3D

Modal Analysis
Linear Static

F1
S1, S12, S13, S14A, S14B, S22, S26, S33A, S33M

COSMOSM Basic FEA System

2-1

Linear Static Analysis

Introduction

This chapter contains verification problems to demonstrate the accuracy of the
Linear Static Analysis module STAR.

List of Linear Static Verification Problems

S1: Pin JointedTruss

2-6

S2: Long Thick-Walled Cylinder

2-7

S3A, S3B: Simply Supported Rectangular Plate

2-9

S4: Thermal Stress Analysis of a Truss Structure

2-10

S5: Thermal Stress Analysis of a 2D Structure

2-12

S6A, S6B: Deflection of a Cantilever Beam

2-13

S7: Beam Stresses and Deflections

2-14

S8: Tip Displacements of a Circular Beam

2-15

S9A: Clamped Beam Subject to Imposed Displacement

2-16

S9B: Clamped Beam Subject to Imposed Rotation

2-18

S10A, S10B: Bending of a Solid Beam

2-19

S11: Thermal Stress Analysis of a 3D Structure

2-21

S12: Deflection of a Hinged Support

2-22

S13: Statically Indeterminate Reaction Force Analysis

2-23

Chapter 2 Linear Static Analysis

2-2

COSMOSM Basic FEA System

List of Linear Static Verification Problems (Continued)

S14A, S14B: Space Truss with Vertical Load

2-24

S15: Out-of-Plane Bending of a Curved Bar

2-25

S16A, S16B: Curved Pipe Deflection

2-26

S17: Rectangular Plate Under Triangular Thermal Loading

2-28

S18: Hemispherical Dome Under Unit Moment Around Free Edge

2-29

S19: Hollow Thick-walled Cylinder Subject to Temperature and

Pressure

2-30

S20A, S20B: Cylindrical Shell Roof

2-31

S21A, S21B: Antisymmetric Cross-Ply Laminated Plate (SHELL4L)

2-33

S22: Thermally Loaded Support Structure

2-35

S23: Thermal Stress Analysis of a Frame

2-36

S24: Thermal Stress Analysis of a Simple Frame

2-38

S25: Torsion of a Square Box Beam

2-39

S26: Beam With Elastic Supports and a Hinge

2-41

S27: Frame Analysis with Combined Loads

2-43

S28: Cantilever Unsymmetric Beam

2-45

S29A, S29B: Square Angle-Ply Composite Plate Under Sinusoidal

2-47

S30: Effect of Transverse Shear on Maximum Deflection

2-48

S31: Square Angle-Ply Composite Plate Under Sinusoidal Loading

2-50

S32A, S32B, S32C, S32D, S32M: Substructure of a Tower

2-51

S33A, S33M: Substructure of an Airplane (Wing)

2-53

S34: Tie Rod with Lateral Loading

2-54

S35A, S35B: Spherical Cap Under Uniform Pressure (Solid)

2-56

S36A, S36B: Substructure of a Simply Supported Plate

2-58

S37: Hyperboloidal Shell Under Uniform Ring Load Around Free

Edge

2-60

S38: Rotating Solid Disk

2-62

S39: Unbalanced Rotating Flywheel

2-63

S40: Truss Structure Subject to a Concentrated Load

2-64

S41: Reactions of a Frame Structure

2-65

COSMOSM Basic FEA System

2-3

Part 2 Verification Problems

List of Linear Static Verification Problems (Continued)

S42A, S42B: Reactions and Deflections of a Cantilever Beam

2-66

S43: Bending of a T Section Beam

2-67

S44A, S44B: Bending of a Circular Plate with a Center Hole

2-68

S45: Eccentric Frame

2-70

S46, S46A, S46B: Bending of a Cantilever Beam

2-71

S47, S47A, S47B: Bending of a Cantilever Beam

2-73

S48: Rotation of a Tank of Fluid (PLANE2D Fluid)

2-75

S49A, S49B: Acceleration of a Tank of Fluid (PLANE2D Fluid)

2-77

S50A, S50B, S50C, S50D, S50F, S50G, S50H, S50I: Deflection

of a Curved Beam

2-79

S51: Gable Frame with Hinged Supports

2-80

S52: Support Reactions for a Beam with Intermediate Forces and

Moments

2-81

S53: Beam Analysis with Intermediate Loads

2-83

S54: Analysis of a Plane Frame with Beam Loads

2-85

S55: Laterally Loaded Tapered Beam

2-86

S56: Circular Plate Under a Concentrated Load (SHELL9 Element)

2-87

S57: Test of a Pinched Cylinder with Diaphragm (SHELL9 Element)

2-89

S58A, S58B, S58C: Deflection of a Twisted Beam with Tip Force

(SHELL9 and TETRA4R Elements)

2-91

S59A, S59B, S59C: Sandwich Square Plate Under Uniform

Loading (SHELL9L)

2-93

S60: Clamped Square Plate Under Uniform Loading

2-95

S61: Single-Edge Cracked Bend Specimen, Evaluation of Stress

Intensity Factor Using Crack Element

2-97

S62: Plate with Central Crack

2-98

S63: Cyclic Symmetry Analysis of a Hexagonal Frame

2-99

S64A, S64B: Cyclic Symmetry

2-100

S65: Fluid-Structure Interaction, Rotation of a Tank of Fluid

2-101

S66: Fluid-Structure Interaction, Acceleration of a Tank of Fluid

2-103

S67: MacNeal-Harder Test

2-105

S68: P-Method Solution of a Square Plate with Hole

2-106

Chapter 2 Linear Static Analysis

2-4

COSMOSM Basic FEA System

List of Linear Static Verification Problems (Concluded)

S69: P-Method Analysis of an Elliptic Membrane Under Pressure

2-107

S70: Thermal Analysis with Temperature Dependent Material

2-108

S71: Sandwich Beam with Concentrated Load

2-109

S74: Constant Stress Patch Test (TETRA4R)

2-110

S75: Analysis of a Cantilever Beam with Gaps, Subject to Different

Loading Conditions

2-111

S76: Simply Supported Beam Subject to Pressure from a Rigid

Parabolic Shaped Piston

2-113

S77: Bending of a Solid Beam Using Direct Material Matrix Input

2-115

S78: P-Adaptive Analysis of a Square Plate with a Circular Hole

2-117

S79: Hemispherical Shell Under Unit Moment Around Free Edge

2-119

S80: Axisymmetric Hyperbolic Shell Under a Cosine Harmonic

Loading on the Free Edge

2-120

S81: Circular Plate Under Non-Axisymmetric Load

2-121

S82: Twisting of a Long Solid Shaft

2-123

S83: Bending of a Long Solid Shaft

2-125

S84: Submodeling of a Plate

2-127

S85: Plate on Elastic Foundation

2-128

S86: Plate with Coupled Degrees of Freedom

2-129

S87: Gravity Loading of ELBOW Element

2-130

S88A, S88B: Single-Edge Cracked Bend Specimen, Evaluation

of Stress Intensity Factor Using the J-integral

2-131

S89A, S89B: Slant-Edge Cracked Plate, Evaluation of Stress

Intensity Factors Using the J-integral

2-133

S90A, S90B: Penny-Shaped Crack in Round Bar, Evaluation of

Stress Intensity Factor Using the J-integral

2-135

S91: Crack Under Thermal Stresses, Evaluation of Stress Intensity

Using the J-integral

2-137

S92A, S92B: Simply Supported Rectangular Plate, Using Direct

Material Matrix Input

2-139

S93: Accelerating Rocket

2-141

S94A, S94B, S94C: P-Method Solution of a Square Plate with Small

Hole

2-143

S95A, S95B, S95C: P-Method Solution of a U-Shaped

Circumferential Groove in Circular Shaft

2-146

COSMOSM Basic FEA System

2-5

Part 2 Verification Problems

S96A, S96B, S96C: P-Method Solution of a Square Plate with an

Elliptical Hole

2-149

Chapter 2 Linear Static Analysis

2-6

COSMOSM Basic FEA System

TYPE:

Static analysis, truss element (TRUSS3D).

REFERENCE:

Beer, F. P., and Johnston, E. R., Jr., “Vector Mechanics for Engineers: Statics and
Dynamics,” McGraw-Hill Book Co., Inc. New York, 1962, p. 47.

PROBLEM:

A 50 lb load is supported by three bars which are attached to a ceiling as shown.
Determine the stress in each bar.

Figure S1-1

S1: Pin Jointed Truss

See

List

GIVEN:

COMPARISON OF RESULTS

Area of each bar = 1 in

E = 30 x 10

psi

1-4

, psi

2-4

, psi

3-4

, psi

Theory

10.40

31.20

22.90

COSMOSM

10.39

31.18

22.91

6 ft

2 ft

6 ft

8 ft

4 ft

Problem Sketch and

Finite Element Model

COSMOSM Basic FEA System

2-7

Part 2 Verification Problems

TYPE:

Static analysis, 2D axisymmetric elements (PLANE2D).

REFERENCE:

Timoshenko, S. P. and Goodier, J., “Theory of Elasticity,” McGraw-Hill, New York,
1951, pp. 58-60.

PROBLEM:

Calculate the radial stresses for an infinitely long, thick walled cylinder subjected to
an internal pressure p.

GIVEN:

= 100 in

b =

115

p =

1000

psi

E = 30 x 10

psi

ν =

0.3

MODELING HINTS:

The model is meshed with three elements through the thickness and three elements
along the length.

COMPARISON OF RESULTS:

S2: Long Thick-Walled Cylinder

See

List

r (Radial Distance)

(in)

Radial Stress

(psi)

Theory

COSMOSM

102.5 (Element 1)

-802.40

-802.51

107.5 (Element 2)

-447.75

-447.84

112.5 (Element 3)

-139.34

-139.42

Chapter 2 Linear Static Analysis

2-8

COSMOSM Basic FEA System

Figure S2-1

1 rad

Finite Element Model

Problem Sketch

COSMOSM Basic FEA System

2-9

Part 2 Verification Problems

TYPE:

Static analysis, 3-node thin plate element (SHELL3).

REFERENCE:

Timoshenko, S. P. and Woinowsky-Krieger, “Theory of Plates and Shells,”
McGraw-Hill Book Co., 2nd edition. pp. 143-120, 1962.

PROBLEM:

Calculate the deflection
at the center of a simply
supported isotropic
plate subjected to (A)
concentrated load F, (B)
uniform pressure (P).

GIVEN:

= 30,000,000 psi

= 0.3

= 1 in

= b = 40 in

= 400 lb

= 1 psi

MODELING HINTS:

Due to double symmetry in geometry and loads, only a quarter of the plate is
modeled.

COMPARISON OF RESULTS:

S3A, S3B: Simply Supported Rectangular Plate

See

List

Case

X (in)

Y (in)

Deflection at Node 25 (UZ)

Theory

COSMOSM

-0.0270230 in

-0.027123 in

-0.00378327 in

-0.0037915 in

Problem Sketch and Finite Element Model

Figure S3-1

Chapter 2 Linear Static Analysis

2-10

COSMOSM Basic FEA System

TYPE:

Linear thermal stress analysis, truss elements (TRUSS2D).

REFERENCE:

Hsieh, Y. Y. “Elementary Theory of Structures,” Prentice-Hall, Inc., 1970, pp. 200-
202.

PROBLEM:

Determine the member forces in truss stucture shown in the figure subject to a 50

F rise in temperature at the top chords (elements 13 and 14).

GIVEN:

E = 30 x 10

psi

Coefficient of thermal expansion =

α = 0.65 x 10

-5

°F

L(ft)/A(in

) = 1for all members

COMPARISON OF RESULTS:

COSMOSM results are calculated by listing element stress results and multiplying
by the corresponding area.

S4: Thermal Stress Analysis of a Truss Structure

See

List

Member Forces (kips

Members

Theory

COSMOSM

Members

Theory

COSMOSM

35.1

-21.1

35.1

-28.1

-21.1

COSMOSM Basic FEA System

2-11

Part 2 Verification Problems

Figure S4-1

4 x @ 24 ft = 96 ft

32 ft

Problem Sketch and Finite Element Model

Chapter 2 Linear Static Analysis

2-12

COSMOSM Basic FEA System

TYPE:

Linear thermal stress analysis, 2D elements (plane strain, PLANE2D).

PROBLEM:

Determine the displacements and stresses of the plane strain problem shown in
figure due to a uniform temperature rise.

Figure S5-1

S5: Thermal Stress Analysis of a 2D Structure

See

List

GIVEN:

= 30 x 10

psi

α = 0.65 x 10

-5

°F

= 0.25

= 100

° F

= 1 in

COMPARISON OF RESULTS:

Displacements at Nodes (2, 4, and 6)

Y-Displacement

(in)

SX-Stress

(psi)

Theory

0.001083

-26000.0

COSMOSM

0.001083

-26000.1

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-13

Part 2 Verification Problems

TYPE:

Static analysis, plane stress element PLANE2D and SHELL6.

PROBLEM:

A cantilever beam is subjected to a concentrated load at the free end. Determine the
deflections at the free end and the average shear stress.

Figure S6-1

S6A, S6B: Deflection of a Cantilever Beam

See

List

GIVEN:

COMPARISON OF RESULTS:

= 30 x 10

psi

= 10 in

h = 1 in
A = 0.1 in

= 0

= 1 lb

* averaged results of
nodes at the free edge

Max. Deflection in

the Y-direction

Shear Stress

(psi)

Theory

-0.001333

-10.0

COSMOSM

PLANE2D

-0.001337

-10.0*

SHELL6

(Curved)

-0.0013398

-9.820667*

SHELL6

(Assembled)

-0.00072411

--8.530667*

Finite Element Model

Problem Sketch

Chapter 2 Linear Static Analysis

2-14

COSMOSM Basic FEA System

TYPE:

Static analysis, beam elements (BEAM3D).

REFERENCE:

Timoshenko, S. P., “Strength of Materials, Part 1, Elementary Theory and
Problems,” 3rd Ed., D. Van Nostrand Co., Inc., New York, 1965, p. 98.

PROBLEM:

A standard 30" Wide Flange beam is supported as shown below and loaded on the
overhangs by a uniformly distributed load of 10,000 lb per ft. Determine the
maximum stress in the middle portion of the beam and the deflection at the center of
the beam.

MODELING HINTS:

Use consistent length units. A half-model has been used because of symmetry.
Resultant force and moment have been applied at node 2 instead of distributed load.

Figure S7-1

S7: Beam Stresses and Deflections

See

List

GIVEN:

Area = 50.65 in

= 30 x 10

psi

= 10,000 lb/ft

COMPARISON OF RESULTS:

At the middle of the span (node 3):

max

psi

inch

Theory

11400.0

0.182

COSMOSM

11400.0

0.182

Finite Element Model

15"

Section a-a

10'

20'

Problem Sketch

COSMOSM Basic FEA System

2-15

Part 2 Verification Problems

TYPE:

Static analysis, thin or thick shell element (SHELL3).

REFERENCE

Warren C. Young, “Roark's Formulas for Stress and Strain,” Sixth Edition, McGraw
Hill Book Company, New York, 1989.

PROBLEM:

Determine the deflections in X, Y direction of a circular beam fixed at one end and
free at the other end, when subjected to a force along X direction at force end.

Figure S8-1

S8: Tip Displacements of a Circular Beam

See

List

GIVEN:

= 30E6 psi

= 0

= 4 in

= 1 in

= 10 in

= 200 lb

COMPARISON OF RESULTS:

The loaded end.

Displacement (inch)

Theory

0.712E-2

0.99E-2

COSMOSM

0.718E-2

0.99E-2

F/2

Problem Sketch and Finite Element Model

Chapter 2 Linear Static Analysis

2-16

COSMOSM Basic FEA System

TYPE:

Static analysis, beam elements (BEAM2D).

REFERENCE

Gere, J. M. and Weaver, W. Jr., “Analysis of Framed Structures,” D. Van Nostrand
Co., 1965.

PROBLEM:

Determine the end forces of a clamped beam due to a 1 inch settlement at the right
end.

GIVEN:

= 30 x 10

psi

= 80 in

A = 4 in

= 1.33 in

= 2 in

ANALYTICAL SOLUTION:

Reaction: R = -12EI / L

Moment: M = 6EI / L

COMPARISON OF RESULTS:

S9A: Clamped Beam Subject

to Imposed Displacement

See

List

Theory

COSMOSM

Imposed Displacement (in)

-1.0

End Shear (lb)

-937.5

End Moment (lb-in)

-37,500.0

COSMOSM Basic FEA System

2-17

Part 2 Verification Problems

Figure S9A-1

1.0 in

Problem Sketch

Finite Element Model

Chapter 2 Linear Static Analysis

2-18

COSMOSM Basic FEA System

TYPE:

Static analysis, beam elements (BEAM2D).

REFERENCE:

Gere, J. M. N. and Weaver, W. Jr., “Analysis of Framed Structures,” D. Van Nostrand
Co., 1965.

PROBLEM:

Determine the end forces of a clamped-clamped beam due to a 1 radian imposed
rotation at the right end.

COMPARISON OF RESULTS:

Figure S9B-1

S9B: Clamped Beam Subject to Imposed Rotation

See

List

GIVEN:

= 30 x 10

psi

= 80 in

A = 4 in

= 1.3333 in

= 2 in

ANALYTICAL SOLUTION:

Reaction: R = -6EI / L

Moment: M = 4EI / L

Theory

COSMOSM

Imposed Rotation (1 rad)

End Shear

-37,500

End Moment

-2,000,000

= 1 rad

Problem Sketch

COSMOSM Basic FEA System

2-19

Part 2 Verification Problems

TYPE:

Static analysis, SOLID element.

REFERENCE:

Roark, R. J., “Formulas for Stress and Strain,” 4th Edition, McGraw-Hill Book Co.,
New York, 1965, pp. 104-106.

PROBLEM:

A beam of length L and height h is built-in at one end and loaded at free end: (A)
with a shear force F, and (B) a moment M. Determine the deflection at the free end.

GIVEN:

= 10 in

= 2 in

= 30 x 10

psi

= 0

= 300 lb

M = 2000 in-lb

MODELING HINTS:

Two load cases have been used (S10A, S10B).

Four forces equal to F/4 have been applied at nodes 21, 22, 23, and 24 in xz
direction (S10A), and,

Two couples equal M/2 have been applied at nodes 21, 22, 23 and 24 (S10B).

COMPARISON OF RESULTS:

Displacement in Z-direction (in) (node 21-24):

S10A, S10B: Bending of a Solid Beam

See

List

S10A

S10B

Theory

0.00500

-0.00500

COSMOSM

0.005007

-0.00495

Chapter 2 Linear Static Analysis

2-20

COSMOSM Basic FEA System

Figure S10A-1

Case 1

Case 2

Problem Sketch

Finite Element Model

COSMOSM Basic FEA System

2-21

Part 2 Verification Problems

TYPE:

Linear thermal stress analysis, 3D SOLID element.

PROBLEM:

Determine the displacements of the three-dimensional structure shown below due to
a uniform temperature rise.

Figure S11-1

S11: Thermal Stress Analysis of a 3D Structure

See

List

GIVEN:

COMPARISON OF RESULTS:

E = 3 x 10

psi

α = 0.65 x 10

-5

°F

= 0.25

T =

100

° F

= 1 in

X-Displacement (Nodes)

5, 6, 7, 8

9, 10, 11, 12

Theory

0.000650

0.001300

COSMOSM

0.000650

0.001300

x,r

z,t

y,s

Problem Sketch and Finite Element Model

Chapter 2 Linear Static Analysis

2-22

COSMOSM Basic FEA System

TYPE:

Static analysis, truss element (TRUSS3D).

REFERENCE:

Timoshenko, S. P., and MacCullough, Glesson, H., “Elements of Strength of
Materials,” D. Van Nostrand Co., Inc., 3rd edition, June 1949, p. 13.

PROBLEM:

A structure consisting of two equal steel bars, 15 feet long and with hinged ends, is
submitted to the action of a vertical load P. Determine the forces in the members AB
and BC along with the vertical deflection at B.

Figure S12-1

S12: Deflection of a Hinged Support

See

List

GIVEN:

= 5000 lbs

θ = 30°
AB = BC = 15 ft
E

= 30 x 10

psi

Cross-sectional area

= 0.5 in

COMPARISON OF RESULTS:

Theory

COSMOSM

Vertical Deflection at
B in inches

0.12

Forces in Members
AB and BC in lbs

5000

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-23

Part 2 Verification Problems

TYPE:

Static analysis, truss elements (TRUSS3D).

REFERENCE:

Timoshenko, S. P., “Strength of Materials, Part 1, Elementary Theory and
Problems,” 3rd edition, D. Van Nostrand Co., Inc., 1956, p. 26.

PROBLEM:

A prismatic bar with built-in ends is loaded axially at two intermediate cross-
sections by forces F1 and F2. Determine the reaction forces R1 and R2.

Figure S13-1

S13: Statically Indeterminate

Reaction Force Analysis

See

List

GIVEN:

COMPARISON OF RESULTS:

= b = 0.3 L

L =

= 2F

= 1000 lb

E = 30 x 10

psi

lbs

Theory

900

600

COSMOSM

900

600

Finite Element Model

Problem Sketch

Chapter 2 Linear Static Analysis

2-24

COSMOSM Basic FEA System

TYPE:

Static analysis, truss elements (TRUSS3D).

REFERENCE:

Timoshenko, S. P. and Young, D. H. “Theory of Structures,” end Ed., McGraw-Hill,
New York, 1965, pp. 330-331.

PROBLEM:

The simple space truss shown in the figure below consists of two panels ABCD and
ABEF, attached to a vertical wall at points C, D, E, F, the panel ABCD being in a
horizontal plane. All bars have the same cross-sectional area, A, and the same
modulus of elasticity, E.

Calculate:

The axial force produced in the
redundant bar AD by the vertical
load P = 1 kip at joint A (S14A).

The thermal force induced in the
bar AD if there is a uniform rise
in temperature of 50

° F (S14B).

GIVEN:

E = 30 x 10

psi

= 6.5 x 10

-6

°F

A =

1in

L = 4 ft

COMPARISON OF RESULTS:

For Element 2:

S14A, S14B: Space Truss with Vertical Load

See

List

S14A

S14B

Theory

56.0 lb

-1259.0 lb

COSMOSM

55.92 lb

-1292.4 lb

Figure S14-1

Problem Sketch and Finite

Element Model

COSMOSM Basic FEA System

2-25

Part 2 Verification Problems

TYPE:

Static analysis, curved elbow element (ELBOW).

REFERENCE:

Timoshenko, S. P., “Strength of Materials, Part 1, Advanced Theory and Problems,”
3rd Edition, D. Van Nostrand Company, Inc., New York, 1956, p. 412.

PROBLEM:

A portion of a horizontal circular ring, built-in at A, is loaded by a vertical load P
applied at the end B. The ring has a solid circular cross-section of diameter d.
Determine the deflection at end B, and the maximum bending stress.

MODELING HINTS:

COSMOSM does not yet have a curved beam element, although this element will be
incorporated into the program shortly. Hence, the curved elbow element is used to
model this problem. Therefore, it is necessary to use equivalent thickness t which is
equal to the radius of the solid rod.

Figure S15-1

S15: Out-of-Plane Bending of a Curved Bar

See

List

GIVEN:

COMPARISON OF RESULTS

P =

r =

100

d = 2 in

E = 30 x 10

psi

α = 90°

= 0.3

, inch

Bend

, psi

Theory

-2.648

6366.0

COSMOSM

-2.650

6366.2

Problem Sketch

Finite Element Model

Chapter 2 Linear Static Analysis

2-26

COSMOSM Basic FEA System

TYPE:

Static analysis, elbow element (ELBOW).

REFERENCE:

Blake, A., “Design of Curved Members for Machines,” Industrial Press, New York,
1966.

PROBLEM:

Calculate deflections x and y for a curved pipe shown in the figure subjected to:

Moment Mz = 3 x 10

lb-in and internal pressure p = 900 psi (S16A).

Internal pressure p = 900 psi (S16B).

GIVEN:

E = 30 x 10

psi

0.3

R =

Thickness = 1.031 in
Outer diameter of pipe = 20 in

COMPARISON OF RESULTS:

Blake gives the following results for a 90 curved member. These results do not
include the effects of distortion of the cross-section and internal pressure.

δy = M

/EI = 0.187039 in

δy = M

/EI (P/2-1) = 0.106761 in

The pipe flexibility factor is given by

= 1.65/h{1 + 6P/Eh) (R/t)

4/3

}, where h = tR/r

for p = 900 psi, K

= 1.8814761

To obtain the nodal deflections for case l, the deflections calculated by Blake's
formulas must be multiplied by kp and added to the deflections produced by the
internal pressure.

S16A, S16B: Curved Pipe Deflection

See

List

COSMOSM Basic FEA System

2-27

Part 2 Verification Problems

S16A

S16B

Figure S16A-1

, inch

Theory

0.37035

0.20515

COSMOSM

0.37034

0.20515

, inch

Theory

1.84356 x 10

-2

4.2873 x 10

-3

COSMOSM

1.84355 x 10

-2

4.28043 x 10

-3

M z

R = Constant

Problem Sketch and Finite Element Model

Chapter 2 Linear Static Analysis

2-28

COSMOSM Basic FEA System

TYPE:

Linear thermal stress analysis, 2D elements (plane stress analysis, PLANE2D).

REFERENCE:

Johns, D. J., “Thermal Stress Analysis,” Pergamon Press, Inc., 1965, pp. 40-47.

PROBLEM:

A finite rectangular plate is subjected to a temperature distribution in only one
direction as shown in figure. Determine the normal stress at point A.

GIVEN:

= 15 in

= 10 in

= -100

°F

= 1 in

= 30 x 10

psi

= 0.65 x 10

-5

in/in/

°F

MODELING HINTS:

Due to the double
symmetry in geometry and
loading, only one quarter
of the plate was analyzed.

COMPARISON OF RESULTS:

S17: Rectangular Plate Under

Triangular Thermal Loading

See

List

/ (E

) (Node 45)

Reference

Method 1

0.42

Method 2

0.40

COSMOSM

0.437

Same Boundary Condition

ounda

ondi

Figure S17-1

COSMOSM Basic FEA System

2-29

Part 2 Verification Problems

TYPE:

Static linear analysis, axisymmetric shell element (SHELLAX).

REFERENCE:

Zienkiewicz, O. C. “The Finite Element Method,” Third edition, McGraw-Hill Book
Co., New York, 1983, p. 362.

PROBLEM:

Determine the horizontal displacement of a hemispherical shell under uniform unit
moment around the free edge.

MODELING HINTS:

Nodal spacing is shown in the Figure. For convenience, cylindrical coordinate
system is chosen for node generation. It is important to note that nodal load is to be
specified per unit radian which in this case is 50 in lb/rad.

Figure S18-1

S18: Hemispherical Dome Under

Unit Moment Around Free Edge

See

List

GIVEN:

= 100 in

= 50 in

E = 1 x 10

psi

ν =

0.33

t =

M = 1 in lb

COMPARISON OF RESULTS:

Horizontal Displacement

(Node 29) (inch)

Reference

1.580 E-5

COSMOSM

1.589 E-5

Nodal Spacing

14 @ 0.5

interval

7 @ 0.5

interval

3 @ 2.0

interval

4 @ 10.0

interval

Finite Element Model

Problem Sketch

Chapter 2 Linear Static Analysis

2-30

COSMOSM Basic FEA System

TYPE:

Static analysis, 2D axisymmetric element (PLANE2D).

REFERENCE:

Timoshenko, S. P. and Goodier, “Theory of Elasticity,” McGraw-Hill Book Co.,
New York, 1961, pp. 448-449.

PROBLEM:

The hollow cylinder in plane strain is subjected to two independent load conditions.

An internal pressure.

A steady state axisymmetric temperature distribution given by the equation:

T(r) = (Ta/ln(b/a)) · ln(b/r) where Ta is the temperature of the inner surface

and T(r) is the temperature at any radius.

S19: Hollow Thick-walled Cylinder Subject

to Temperature and Pressure

See

List

GIVEN:

E = 30 x 10

psi

= 1 in

b = 2 in

ν =

0.3

α = 1 x 10

-6

in/(in

°F)

Pa = 100 psi
Ta =

100

°F

COMPARISON OF RESULTS:

At r = 1.2875 in (elements 13, 15)

, psi

θ, psi

Theory)

-398.34

-592.47

COSMOSM

-398.15

-596.46

Figure S19-1

T(r)

0.1

Problem Sketch

Finite Element Model

COSMOSM Basic FEA System

2-31

Part 2 Verification Problems

TYPE:

Static analysis, shell element (SHELL4, SHELL6).

REFERENCE:

Pawsley, S. F., “The Analysis of Moderately Thick to Thin Shells by the Finite
Element Method,” Report No. USCEM 70-l2, Dept. of Civil Engineering,
University of California, l970.

PROBLEM:

Determine the vertical deflections across the midspan of a shell roof under its own
weight. Dimensions and boundary conditions are shown in the figure below.

GIVEN:

r =

= 3 x 10

psi

ν =

Shell Weight = 90 lbs/sq ft

MODELING HINTS:

Due to symmetry, a quarter of the shell is considered for modeling. The distributed
force (self weight) is lumped at the nodes.

COMPARISON OF RESULTS:

Vertical Deflection at Midspan of free edge (Node 25):

S20A, S20B: Cylindrical Shell Roof

See

List

, (inch)

Theory

-0.3024

COSMOSM

SHELL4

-0.3036

SHELL6

(Curved)

-0.24580

SHELL6

(Assembled)

-0.29353

Chapter 2 Linear Static Analysis

2-32

COSMOSM Basic FEA System

Figure S20-1

Figure S20-2

25 ft

Free Edge

v =
w = 0

t = 0.25 ft

v = w = 0

Problem Sketch and Finite

Element Model

0.18961

0.30365

-.2

0.07423

0.01335

0.004676

-.1

-.3

-.1

-.2

-.3

COSMOS/M

EXACT

10 15

COSMOSM Basic FEA System

2-33

Part 2 Verification Problems

TYPE:

Static analysis, composite shell element (SHELL4L, SHELL9L).

REFERENCE:

Jones, Robert M., “Mechanics of Composite Materials,” McGraw-Hill, New York,
l975, p. 256.

PROBLEM:

Calculate the maximum deflection of a simply supported antisymmetric cross-ply
laminated plate under sinusoidal load. The plate is made up of 6-layers and the
material in each layer is orthotropic.

GIVEN:

a =

l00

b =

h = l in
E

= 40E6 psi

= lE6 psi

= 0.25

= G

= 5E5 psi

For each layer, pressure loading = cos

π x/a · cos π y/b

MODELING HINT:

Due to symmetry, a quarter of the plate is considered for modeling.

S21A, S21B: Antisymmetric Cross-Ply

Laminated Plate (SHELL4L)

See

List

Chapter 2 Linear Static Analysis

2-34

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

Figure S21-1

Maximum

Deflection (in)

Theory

0.105E-2

COSMOSM

4-nod shell

0.104E-2

9-node shell

0.111E-2

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-35

Part 2 Verification Problems

TYPE:

Static, thermal stress analysis, truss and beam elements (TRUSS3D, BEAM3D).

REFERENCE:

Timoshenko, S. P., “Strength of Materials, Part l, Elementary Theory and Problems,”
3rd Ed., D. Van Nostrand Co., Inc., l956, p. 30.

PROBLEM:

Find the stresses in the copper and steel wire structure shown below. The structure
is subjected to a load Q and a temperature rise of l0

° F after assembly.

MODELING HINTS:

Length and spacing between wires are arbitrarily selected. Truss element is used for
elements number (l), (2), and (3), and the beam element for elements (4) and (5).
Beam type and material are arbitrarily selected.

Figure S22-1

S22: Thermally Loaded Support Structure

See

List

GIVEN:

Cross-sections area = 0.l in

Q = 4000 lb

= 92 x l0 in/in -

°F

= 70 x l0 in/in -

°F

= l6 x l0

psi

= 30 x l0

psi

COMPARISON OF RESULTS:

steel

, psi

copper

, psi

Theory

19695.0

10152.0

COSMOSM

19704.2

10147.9

copper

steel

copper

Problem Sketch

Finite Element Model

RIGID BEAM

20"

10"

Chapter 2 Linear Static Analysis

2-36

COSMOSM Basic FEA System

TYPE:

Linear thermal stress analysis, beam elements (BEAM3D).

REFERENCE:

Rygol, J., “Structural Analysis by Direct Moment Distribution,” Gordon and Breach
Science Publishers, New York, l968, pp. 292-294.

PROBLEM:

S23: Thermal Stress Analysis of a Frame

See

List

An irregular
frame subjected
to differential
temperature.
Find member
end moments.

Member Specifications

Member

d (ft)

b (ft)

Ar-r (ft)

lt-t (ft)

1.5

2.25

0.422

2.25

1.25

2.8125

1.187

2.0

1.5

3.0

1.0

2.5

1.25

3.125

1.628

2.0

1.5

3.0

1.0

GIVEN:

= 192857 tons/ft

α = 0.0000l ft/ft°C

COMPARISON OF RESULTS:

Moments (lb-in):

Member No.

COSMOSM

Reference

Solution

-17.96

+17.96

-42.87

+17.96

-42.96

+38.73

-41.92

+38.64

-41.96

+84.79

-82.61

+84.92

-82.61

-57.50

+82.61

-57.40

+82.61

COSMOSM Basic FEA System

2-37

Part 2 Verification Problems

Figure S23-1

Figure S23-2

18'

27'

12'

40 C

80 C

10 C

Problem Sketch and Finite Element Model

Section A-A

Section B-B

Chapter 2 Linear Static Analysis

2-38

COSMOSM Basic FEA System

TYPE:

Linear thermal stress analysis, beam elements (BEAM2D).

PROBLEM:

Determine displacements and end forces of the frame shown in the figure below due
to temperature rise at the nodes and thermal gradients of members as specified
below.

COMPARISON OF RESULTS:

Displacements at node 2 (in):

Figure S24-1

S24: Thermal Stress Analysis of a Simple Frame

See

List

GIVEN:

= 30,000 kips/in

= 0.65 x l0 in/in/°F

Element

No.

Difference in Temperature (

S-dir

T-dir

72 0

13.5

Theory

-0.0583

0.1157

COSMOSM

-0.0583

0.1168

ctttttiiiiio

100

240"

120"

Problem Sketch and Finite Element Model

width
= 5"

depth

ctttttiiiiio

(y)

(z)

t (z)

s(y)

width
= 3"

depth
= 6"

COSMOSM Basic FEA System

2-39

Part 2 Verification Problems

TYPE:

Static analysis, shell elements (SHELL4).

REFERENCE:

Timoshenko, S. P., and Goodier, J. N., “Theory of Elasticity,” McGraw-Hill, New
York, 1951, p. 299.

PROBLEM:

Find the shear stress and the angle of twist for the square box beam subjected to a
torsional moment T.

GIVEN:

= 7.5 psi

= 0.3

= 3 in

= 150 in

= 1500 in

T = 300 lb in

COMPARISON OF RESULTS:

is calculated as:

θ =

Sin

-1

(resultant displacemet of node 25/distance from node 25 to the center of the cross

section)

S25: Torsion of a Square Box Beam

See

List

Shear Stress

psi

Rotation

, rad

Theory

0.00222

0.0154074

COSMOSM

0.0021337 (average)

0.0154035*

Chapter 2 Linear Static Analysis

2-40

COSMOSM Basic FEA System

Figure S25-1

Problem Sketch

Section

I-I

150

1500

.25

150

Finite Element Model

COSMOSM Basic FEA System

2-41

Part 2 Verification Problems

TYPE:

Static analysis, beam and truss elements (TRUSS3D, BEAM3D).

REFERENCE:

Beaufait, F. W., et. al., “Computer Methods of Structural Analysis,” Prentice-Hall,
Inc., New Jersey, l970, pp. l97-2l0.

PROBLEM:

The final end actions of the members and the reactions of the supports resulting from
the applied loading are to be determined for the structural system described in the
figure below. At the beam-column connection, joint 3, the beam is continuous and
the column is pin-connected to the beam.

GIVEN:

Cross-sectional area of beams

= A

= 0.125 ft

Moment of inertia of beams

= I

= 0.263 ft

Cross-sectional area of column = A

= 0.175 ft

Moment of inertia of column

= I

= 0.193 ft

= 1.44 x 10

kip/ft

K (spring stiffness)

= l200 kips/ft

COMPARISON OF RESULTS:

S26: Beam With Elastic Supports and a Hinge

See

List

Node 2

Node 4

Node 5

Reference

(10

-3

ft)

(10

-3

ft)

(10

-3

rad)

1.0787

1.7873

0.0992

1.0787

-4.8205

0.3615

1.0787

-0.1803

-0.4443

COSMOSM

(10

-3

ft)

(10

-3

ft)

(10

-3

rad)

1.0794

1.7869

0.0992

1.0794

-4.8205

0.3615

1.0794

-0.1803

-0.4443

Chapter 2 Linear Static Analysis

2-42

COSMOSM Basic FEA System

Figure S26-1

Figure S26-2

12'

24'

30'

15k

Problem Sketch and Finite Element Model

15k

2.1k
(2.144)

51.4 ft-k
(51.46)

9.2 k
(9.215)

5.8k
(5.785)

11.3k
(11.36)

5k
(5)

5.8k
(5.785)

30.0 ft-k
(30.0)

COSMOSM Basic FEA System

2-43

Part 2 Verification Problems

TYPE:

Static analysis, beam elements (BEAM3D).

REFERENCE:

Laursen, Harold I., “Structural Analysis,” McGraw-Hill Book Co., Inc., New York,
1969, pp. 310-312.

PROBLEM:

Determine the forces in the beam members under the loads shown in the figure.
Consider two separate load cases represented by the uniform pressure and the
concentrated force. Set up the input to solve each one individually and then combine
them together to obtain the final result.

GIVEN:

= I

= 0.3215 ft

= 0.6430 ft

= 3.50 ft

2,3

= 4.40 ft

= 2.79 ft

= 432 x 10

K/ft

Areas of members were made to be larger than the actual area in order to neglect
axial deformation.

COMPARISON OF RESULTS:

The results are shown in the figure below with COSMOSM results shown in
parentheses.

S27: Frame Analysis with Combined Loads

See

List

Chapter 2 Linear Static Analysis

2-44

COSMOSM Basic FEA System

Figure S27-1

Figure S27-2

15'

0.5 K/ft

E, I

Problem Sketch

Finite Element Model

10.547 K ft
(10.51)

6.766K ft
(6.76)

28.256K ft
(28.32)

10.682 K ft
(10.67)

COSMOSM Basic FEA System

2-45

Part 2 Verification Problems

TYPE:

Static analysis, 3D beam element (BEAM3D).

REFERENCE:

Boresi, A. P., Sidebottom, O. M., Seely, F. B., Smith, J. O., “Advanced Mechanics
of Materials,” John Wiley and Son, Third Edition, 1978.

PROBLEM:

An unsymmetric cantilever beam is subjected to a concentrated load at the free end.
Determine the tip displacement of the beam, the end forces and the stress at y = 8,
z = -2 at the clamped end.

GIVEN:

COMPARISON OF RESULTS:

S28: Cantilever Unsymmetric Beam

See

List

= 2 x 10

N/cm

Fy = -8 N
h

= 4 cm

= 2 cm

= 500 cm

= -4 N

= 8 cm

= 6 cm

A = 19 cm

= 100.3 cm

= 278.3 cm

= 97.3 cm

= J = 6.333 cm

= 1 cm

Theory

COSMOSM

Node 6

Translation in Y Dir (cm)
Translation in Z Dir (cm)
Rotation about X Axis (rad)

-0.1347
-0.2140

-0.1346

-0.21364

Node 1

Moment about Y Axis (N-cm)
Shear in Z Dir (N)
Stress at Y = 8, Z = 2

-2000.0

4.0

155.74 Tension

-2000.0

4.0

155.8 Tension

Chapter 2 Linear Static Analysis

2-46

COSMOSM Basic FEA System

Figure S28-1

S.C.

Unsymmetric Beam Structure

C.G.

Cross Section

Problem Sketch

Finite Element Model

COSMOSM Basic FEA System

2-47

Part 2 Verification Problems

TYPE:

Static analysis, composite shell (SHELL9L), and solid element (SOLIDL).

REFERENCE:

Jones, Robert M., “Mechanics of Composite Materials,” McGraw-Hill, N. Y., l975,
p. 258.

PROBLEM:

Calculate the maximum deflection of a simply supported square antisymmetric
angle-ply under SINUSOIDAL loading. The plate is made up of 6 layers, where the
top layer material axis orientation makes 45 degree angle with x-axis. To impose
simply-supported boundary conditions, 2 layers of composite solid elements (each
has 3 layers of different
material orientation)
through the thickness
are required.

GIVEN:

= b = 20 in

40E6

psi

= 0.01 in

= lE6 psi

ν =

0.25

= G

= 5E5 psi

= cos (

π x/a) cos (π y/b)

= 1E-3

COMPARISON OF RESULTS:

S29A, S29B: Square Angle-Ply Composite

Plate Under Sinusoidal Loading

See

List

Max. Deflection

Reference Solution

0.256

COSMOSM

SOLIDL

0.258

SHELL9L

0.258

Problem Sketch and

Finite Element Model

Figure S29-1

Chapter 2 Linear Static Analysis

2-48

COSMOSM Basic FEA System

TYPE:

Static analysis, shell elements (SHELL3).

REFERENCE:

Pryor, Charles W., Jr., and Barker, R. M., “Finite Element Bending Analysis of
Reissner Plates,” Engineering Mechanics Division, ASCE, EM6, December, l970,
pp. 967-983.

PROBLEM:

Find the effect of transverse shear on maximum deflection of an isotropic simply
supported plate subjected to a constant pressure, q.

GIVEN:

MODELING HINTS:

The input data corresponds to h = 0.1008 and the other inputs can be obtained by
changing the thickness in the given input data. Due to symmetry, only one quarter of
the plate is considered.

COMPARISON OF RESULTS:

S30: Effect of Transverse Shear

on Maximum Deflection

See

List

= b = 24 in

= 30E6 psi

H = varies according to

thickness ratio (H/a)

= 0.3

= 30 psi

Thickness

Ratio H/a

Thickness H

Coefficient

Difference

(%)

Reissner Theory

COSMOSM

0.000042

0.001008

0.04436

0.04438

0.045

0.00042

0.01008

0.04436

0.04438

0.045

0.0042

0.1008

0.04436

0.04438

0.045

0.05

1.20

0.044936

0.044772

***

0.36

0.1

2.40

0.046659

0.046510

***

0.32

0.15

3.60

0.049533

0.049405

***

0.26

0.2

4.80

0.053555

0.053458

***

0.180

0.25

6.00

0.058727

0.058669

***

0.10

0.3

7.20

0.065048

0.065038

***

0.02

= EH

max

/qa

Thin Shell (SHELL3)

Thick

Shell (SHELL3T)

COSMOSM Basic FEA System

2-49

Part 2 Verification Problems

Figure S30-1

Figure S30-2

Problem Sketch

6.0

Present Finite Element

Reissner's Theory

Classical Theory

0.0

0.3

0.05

1.0

Thickness Ratio, H/a

Coefficient,

x10

b =

max

0.0

Chapter 2 Linear Static Analysis

2-50

COSMOSM Basic FEA System

TYPE:

Static analysis, shell element (SHELL4L).

REFERENCE:

Jones, Robert M., “Mechanics of Composite Materials,” McGraw Hill, N. Y., l975,
p. 258.

PROBLEM:

Calculate the maximum deflection of a simply supported square antisymmetric
angle-ply under sinusoidal loading. The plate is made of 4-layers where the top layer
material axis orientation makes l5 degree angle with the X-axis.

GIVEN:

= b = 20 in

= 1 in

= 40E6 psi

= 1E6 psi

= 0.25

= G

= 5E5 psi

= cos (

π x/a)

cos (

π y/b)

COMPARISON OF RESULTS:

S31: Square Angle-Ply Composite

Plate Under Sinusoidal Loading

See

List

max

(inch)

Theory

4.24E-4

COSMOSM

4.40E-4

Figure S31-1

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-51

Part 2 Verification Problems

TYPE:

Static analysis, substructuring using truss elements (TRUSS2D).

PROBLEM:

Determine the deflections of a tower loaded at top, using multi-level substructures.

GIVEN:

COMPARISON OF RESULTS:

S32A, S32B, S32C, S32D, S32M:

Substructure of a Tower

See

List

= 10 x l0

psi

= 1000 lb

= l00 in

= 30 in

Cross-sectional areas of vertical
and horizontal bars = l in

Cross-sectional areas of diagonal
bars = 0.707 in

Node Number

Deflection (10

-3

inch)

Full

Structure

Sub-

structure

COSMOSM Using

Full Structure

COSMOSM Using

Substructure

1(A)

5(A)

7.9761

9.1679

7.9761

9.1678

5(B)

14.6226

25.3632

14.6226

25.3631

5(C)

19.9360

46.8896

19.9359

46.8893

5(M)

23.9168

71.9909

23.9167

71.9905

3(D)

26.5878

99.3979

26.5877

99.3973

4(A)

-2.1815

3.4329

-2.1815

3.4328

2(B)

-4.0240

8.9940

-4.0239

8.9940

2(C)

-6.7108

25.1829

-6.7108

25.1828

2(M)

-8.0642

46.7075

-8.0641

46.7072

6(M)

-8.0834

71.7683

-8.0833

71.7678

4(D)

-6.7458

98.0790

-6.7457

98.0784

Chapter 2 Linear Static Analysis

2-52

COSMOSM Basic FEA System

Figure S32A-1

Problem Sketch

(Full Structure)

(D) Super Element # 4 - Level

(B) Super Element #2 - Level 2

(M) Main Element

Finite Element Models

Super Nodes

(A) Super Element #1 - Level 3

COSMOSM Basic FEA System

2-53

Part 2 Verification Problems

TYPE:

Static analysis, substructuring using shell, beam and truss elements (SHELL4,
BEAM3D, TRUSS3D).

PROBLEM:

By using substructure method, determine the deflection of an airplane through the
assembly of the calculations concerning separate parts, of the plane.

COMPARISON OF RESULTS:

Figure S33A-1

S33A, S33M: Substructure of an Airplane (Wing)

See

List

Node

Number

Deflection (Z-Direction) (inch)

COSMOSM Using

Full Structure

COSMOSM Using

Substructure

10.100

10.178

8.0671

8.1024

13.800

13.894

10.666

10.708

8.7693

8.9513

12.276

12.958

16'

60'

Problem Sketch

(Full Structure)

Note: Nodes
with are
Super Nodes

Chapter 2 Linear Static Analysis

2-54

COSMOSM Basic FEA System

TYPE:

Static analysis, stress stiffening, beam elements (BEAM3D).

REFERENCE:

Timoshenko, S. P., “Strength of Materials, Part II, Advanced Theory and Problems,”
3rd Edition, D. Von Nostrand Co., Inc., New York, l956, p.42.

PROBLEM:

A tie rod subjected to the action of a tensile force S and a uniform lateral load q.
Determine the maximum deflection z, and the slope at the left end. In addition,
determine the same two quantities for the unstiffened tie rod (S = 0).

GIVEN:

= 200 in

= 30E6 psi

= 2l,972.6 lb

= l.79253 lb/in

= h = 2.5 in

CALCULATED INPUT:

Area = 6.25 in

l =

3.2552

MODELING HINTS:

Due to symmetry, only one-half of the beam is modeled.

S34: Tie Rod with Lateral Loading

See

List

COSMOSM Basic FEA System

2-55

Part 2 Verification Problems

COMPARISON OF RESULTS:

≠

0 (Stiffened)

S = 0 (Unstiffened)

Figure S34-1

max

, in

rad

Theory

-0.2

0.0032352

COSMOSM

-0.19701

0.0031776

max

, in

rad

Theory

-0.382406

0.006115

COSMOSM

-0.37763

0.0060229

Zmax

Problem Sketch

L/2

Finite Element Model

Chapter 2 Linear Static Analysis

2-56

COSMOSM Basic FEA System

TYPE:

Static analysis, solid and composite solid elements (SOLID, SOLIDL).

REFERENCE:

Reddy, N. J. “Exact Solutions of Moderately Thick Laminated Shells,” J. Eng.
Mech. Div. ASCE, Vol. 110, (1984), pp. 794-809.

PROBLEM:

Calculate the center deflection of a simply supported spherical cap under uniform
pressure (q = 1.) in the direction normal to the cap surface. To impose simply-
supported boundary conditions by solid elements, 2 layers of elements through the
thickness are required.

Two types of material properties are being tested, each by a different solid element.

A. Isotropic material is handled by SOLID element (S35A).
B. Composite material, 4 layers with the orientation 0

°/90°/90°/0°, is analyzed by

SOLIDL element (S35B). The lower layer of element is modeled by 2 layers of
material orientation 0

°/90° and the upper one is by 90°/0°.

To capture the geometry of a curved surface by a bi-linear shape function accurately,
at least 8 elements per side have to be used. The model used below is an 8x8x2 mesh.

GIVEN:

S35A, S35B: Spherical Cap Under

Uniform Pressure (Solid)

See

List

Geometry:

= 96

= 0.32 in

Length of side a =b = c = d = 32 in

Material Properties:

1. S35A: Isotropic

E = 1E7 psi
ν = 0.3

2. S35B: Composite 0

°/90°/90°/0°

= 25E6 psi

= E

= 1E6 psi

= 0.25

= 0

= 0.2E6 psi

= G

= 0.5E6 psi

COSMOSM Basic FEA System

2-57

Part 2 Verification Problems

MODELING HINTS:

Boundary Conditions

Due to symmetry:

1. All nodes on plane A, Uy = 0
2. All nodes on plane B, Ux = 0

Simply supported:

1. All nodes on side C, radial displacement

= 0, Disp. on plane C = 0

2. All nodes on side D, radial displacement

= 0, Disp. on plane D = 0

COMPARISON OF RESULTS:

Element

max

(inch)

(1) Isotropic

Exact

SOLID

0.3139E-2
0.3232E-2

(2) Composite

Exact

SOLIDL

0.199E-1

0.2007E-1

243

sym.

162

PLAN

E C

PLAN

E D

PLAN

E A

PLAN

E D

PLAN

E C

Problem Sketch and Finite Element Model

PLAN

E B

163

171

sym

154

235

Figure S35A-1

Chapter 2 Linear Static Analysis

2-58

COSMOSM Basic FEA System

TYPE:

Static analysis, substructuring using plate elements (SHELL4).

REFERENCE:

Timoshenko, S., “Theory of Plates and Shells,” McGraw-Hill, New York, 1940, p.
113-198.

PROBLEM:

Calculate the deflections of a simply supported isotropic plate subjected to uniform
pressure p using the substructuring technique. Nodes 11 through 15 are super nodes
which connect the substructure to the main structure.

GIVEN:

= 30 x 10 psi

= 0.3

= 0.5 in

= 5 psi

= 16 in

= 10 in

MODELING HINTS:

Due to symmetry, a quarter of a plate is taken for modeling.

COMPARISON OF RESULTS:

Timoshenko gives the expression for deflection w in the z-direction with origin at
the corner of the plate.

S36A, S36M: Substructure of a

Simply Supported Plate

See

List

COSMOSM Basic FEA System

2-59

Part 2 Verification Problems

Figure S36A-1

Node No.

(inch)

Theory

COSMOSM

0.0012103

0.00121329

0.0011338

0.00113636

0.0009043

0.00090247

0.0005150

0.00051044

16 17 18 19 20

6 7 8 9 10

11 1 2 13 14 15

21 22 23 24 25

Problem Sketch

1 2 3 4 5

11 1 2 13 14 15

(A) Main Structure

(B) Substructure

Finite Element Models

Chapter 2 Linear Static Analysis

2-60

COSMOSM Basic FEA System

TYPE:

Static analysis, axisymmetric shell elements (SHELLAX).

REFERENCE:

William Weaver, Jr., and Paul R. Johnston, “Finite Elements for Structural
Analysis,” Prentice-Hall, Inc., l984, p. 275.

PROBLEM:

Determine the horizontal displacement of a hyperboloidal shell under uniform ring
load around free edge.

GIVEN:

Equation of the hyperboloid:

= 0.48 (Y - H

)

+ R

MODELING HINTS:

Nodes at the top of the tower are spaced closely because of the concentrated ring
load. Nodal spacing is as follows:

And it is to be noted that the ring load should be input per radian length, since the
radius at the top of the shell is 865.3323 in, the load is 865.33 kip/rad.

S37: Hyperboloidal Shell Under Uniform

Ring Load Around Free Edge

See

List

= 600 in

= l200 in

H = 2400 in
H

= l500 in

= 8 in

= 3000 kip/sq in

= 0.3

= l kip/in

Nodes

1-11

11-21

21-29

29-39

(in)

150

COSMOSM Basic FEA System

2-61

Part 2 Verification Problems

COMPARISON OF RESULTS:

Figure S37-1

Maximum Displacement

at Node 41 (inch)

Theory

-0.904

COSMOSM

-0.89705

Simply
Supported

Problem Sketch

4 1

2 2

2 3

∆

Section

Chapter 2 Linear Static Analysis

2-62

COSMOSM Basic FEA System

TYPE:

Static analysis, axisymmetric (PLANE2D) elements, centrifugal loading.

REFERENCE:

S. P. Timoshenko and J. N. Goodier, “Theory of Elasticity,” McGraw-Hill, New
York, l970, p. 80.

PROBLEM:

A solid disk rotates about center 0 with angular velocity

ω. Determine the stress

distribution in the disk.

GIVEN:

COMPARISON OF RESULTS:

Figure S38-1

S38: Rotating Solid Disk

See

List

= 30 x l0

psi

DENS = 0.02 lb sec

/in

= 0.3

= l in

ω = 25 rad/sec
R

= 9 in

Location

Element 1 (r = 0.5 inch)

Location

Element 9 (r = 8.5 inch)

psi

θ psi

psi

θ psi

Theory

416.37

416.91

45.12

203.16

COSMOSM

416.82

46.18

202.03

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-63

Part 2 Verification Problems

TYPE:

Static analysis, centrifugal loading, beam and mass elements (BEAM3D, MASS).

PROBLEM:

The model shown in the figure is assumed to be rotating about the y-axis at a
constant angular velocity of 25 rad/sec. Determine the axial forces and bending
moments in the supporting beams and columns due to self-weight and rotational
inertia.

GIVEN:

COMPARISON OF RESULTS:

S39: Unbalanced Rotating Flywheel

See

List

Element Properties:

A = 100 in

Iyy = Izz = l000 in

Ixx = 2000 in

= 30 x l0

psi

G = 10 x l0

psi

= 0.0l lb-sec

/in

Inertial Properties:

= m

= l0

= -l00 in/sec

w = 25 rad/sec

Theory

COSMOSM

Element 2, Node 2
Axial Force
Bending Moment

58,800

412,000

58,800

412,000

Element 3, Node 2
Axial Force
Bending Moment

61,200

388,000

61,200

388,000

Figure S39-1

Problem Sketch and Finite Element Model

Chapter 2 Linear Static Analysis

2-64

COSMOSM Basic FEA System

TYPE:

Static analysis, truss elements (TRUSS2D).

REFERENCE:

Hsieh, Y. Y., “Elementary Theory of Structures,” Prentice-Hall Inc., l970, pp. l62-
l63.

PROBLEM:

Calculate the reactions and the vertical deflection of joint 2 of the loaded truss

shown below subject to a concentrated load.

Figure S40-1

S40: Truss Structure Subject to a

Concentrated Load

See

List

GIVEN:

COMPARISON OF RESULTS:

= 30,000 kips/in

= 64 kips

L (ft)/A(in) = 1 for all members

Theory

COSMOSM

Deflection of Joint 2 0.006733 in

0.006733 in

Reaction at Node 1

48 K

Reaction at Node 5

16 K

32 ft

4 at 24 ft = 96 ft

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-65

Part 2 Verification Problems

TYPE:

Static analysis, beam element (BEAM2D).

REFERENCE:

Hsieh, Y. Y., “Elementary Theory of Structures,” Prentice-Hall Inc., l970, pp. 258-
259.

PROBLEM:

Determine the reactions for
the frame shown below.

GIVEN:

= 30 x 10

psi

A = 0.1 in

The relative values of 2EI/L:
for element l = l lb-in
for elements 2, 3 = 2 lb-in

COMPARISON OF RESULTS:

The free body diagram for the structural system is given below and COSMOSM
results are given in parentheses.

Figure S41-2

S41: Reactions of a Frame Structure

See

List

25 ft.

15 ft.

Problem Sketch and Finite Element Model

Figure S41-1

64.28
(64.28)

100

35.72
(35.72)

40.71
(40.71)

343.0
(342.8)

k-ft

40.71
(40.71)

257.2
(257.3)

k-ft

Chapter 2 Linear Static Analysis

2-66

COSMOSM Basic FEA System

TYPE:

Static analysis, shell elements (SHELL4, SHELL6).

PROBLEM:

Calculate reactions and deflections of a cantilever beam subject to a concentrated

load at tip.

Figure S42-1

S42A, S42B: Reactions and

Deflections of a Cantilever Beam

See

List

GIVEN:

COMPARISON OF RESULTS:

= 30E6 psi

= 1 in

= 10 in

W = 4 in
P

= 8 lb

Theory

COSMOSM

SHELL4

SHELL6

(Curved)

SHELL6

(Assembled)

Tip Deflection (Node 33) -2.667 x 10

-4

-2.667 x 10

-4

-2.683 x 10

-4

-2.667 x 10

-4

Total Force Reaction

8 lb

Total Moment Reaction

-80 lb-in

Problem Sketch

COSMOSM Basic FEA System

2-67

Part 2 Verification Problems

TYPE:

Static analysis, shell element, beam element with offset (SHELL4L, BEAM3D).

PROBLEM:

Calculate the deflections and stresses of a cantilever T beam subjected to a
concentrated load at the free end.

Figure S43-1

S43: Bending of a T Section Beam

See

List

GIVEN:

COMPARISON OF RESULTS:

= 2000 in

= 49 in

= 480833.33 in

= 10E10 psi

Dy = -24 in

Theory

COSMOSM

Free End (at Node 12)
Y-Displacement (in)

- Rotation

-5.546E-6
-4.159E-9

-5.588E-6
-4.161E-9

Clamped End

top (psi)

bottom (psi)

4.57360
20.3813

4.377

21.302

ANALYTICAL SOLUTION:

δ = PL

/ 3EI

φ = PL

/ 2EI

σ = Mc / I

NOTE:

The maximum stress occurs in the beam. The
point at which stresses are calculated for
unsymmetric beams should be specified in the
real constant set (real constants 25 and 26).

200"

50"

10"

SHELL4L

C.G. of

BEAM 3D

Element

N.A.

Finite Element Model

Problem Sketch

Chapter 2 Linear Static Analysis

2-68

COSMOSM Basic FEA System

TYPE:

Static analysis, shell elements (SHELL4), coupled points (file S44A) and/or
constraint equations (file S44B).

REFERENCE:

Timoshenko, S., “Strength of Materials, Part II, Advanced Theory and Problems,”
3rd Edition, D. Van Nostrand Co., Inc., New York, l956.

PROBLEM:

A circular plate with a center hole is built-in along the inner edge and unsupported
along the outer edge. The plate is subjected to bending by a moment M applied along
the outer edge. Determine the maximum deflection and the maximum slope of the
plate. In addition, deter-mine the moment M and the corresponding stress at the
center of the first and the last elements.

GIVEN:

CALCULATED INPUT:

= 10 in-lb/in = 52.359 in lb/l0

° segment

MODELING HINTS:

Since the problem is axisymmetric, only a small sector of elements is needed. A
small angle

θ is used for approximating the circular boundary with a straight-side

element. A radial grid with nonuniform spacing (3:l) is used. The load is applied
equally to the outer nodes. Coupled nodes (CPDOF) and/or constraint equations
(CEQN) are used to ensure symmetry for S44A and S44B, respectively. Note that all
constraint and load commands are active in the cylindrical coordinate system.

S44A, S44B: Bending of a Circular

Plate with a Center Hole

See

List

= 30E6 psi

= 0.3

= 0.25 in

= 10 in

= 30 in

M = 10 in lb/in
θ

= 10

COSMOSM Basic FEA System

2-69

Part 2 Verification Problems

COMPARISON OF RESULTS

At the outer edge (node 14).

Figure S44-1

, inch

, rad

Theory

0.0490577

-0.0045089

COSMOSM

0.0492188

-0.0044562

Difference

0.3%

1.17%

The above results are tabulated for S44A.
Identical results will be obtained for S44B.

X = 10.86 inch

(First Element)

X = 27.2 inch

(Sixth Element)

Moment

in-lb/in

, psi

Moment

in-lb/in

, psi

Theory

-13.7

1319

-10.1

971.7

COSMOSM

-13.7

1313

-10.1

972.7

Difference

0.45%

0.10%

The above results are tabulated for S44A. Identical
results will be obtained for S44B.

Problem Sketch

Finite Element Model

)

Chapter 2 Linear Static Analysis

2-70

COSMOSM Basic FEA System

TYPE:

Static analysis, beam elements
(BEAM3D) and point-to-point
constraints (CPCNS
command).

REFERENCE:

Laursen H. I., “Structural
Analysis,” McGraw-Hill, l969.

PROBLEM:

Two vertical beams constitute
an eccentric portal frame with
the aid of 3 horizontal rigid
bars. Find the deformations
resulting from the horizontal
forces.

GIVEN:

= l in

= 2.5

W = l in

E = lE6 psi

= l0 in

= 1 lb

= l/l2 in

(about y and z axis)

MODELING HINT:

Point-to-point constraint elements are used (i.e., points 2-
3, 3-4 and 4-5) to ensure the frame 2-3 - 4-5 is rigid when
the horizontal forces are loaded. Each of the beams (l)
and (2) will deform as shown:

COMPARISON OF RESULTS:

S45: Eccentric Frame

See

List

Deflection
along X-axis
at points A
and B:

(inch)

Theory

1.0000E-3

COSMOSM

1.0104E-3

Difference

Cross Section

Problem Sketch and Finite Element Model

Figure S45-1

Figure S45-2

Deflected S hape

COSMOSM Basic FEA System

2-71

Part 2 Verification Problems

TYPE:

Static analysis, plane stress elements (PLANE2D), beam elements (BEAM2D),
shell elements (SHELL9) and constraint elements.

PROBLEM:

Calculate the maximum deflection and the maximum rotation of a cantilever beam
loaded by a shear force at the free end.

GIVEN:

MODELING HINTS: Continuum-to-Structure Constraint

Problem 1 (S46):

The plane stress elements are defined by nodes 1 through 12. The beam element is
defined by nodes 13 and 14. Each plane stress element is theoretically equivalent to
a beam element where I = 1/12 in

. Node 14 is attached to line 11-12, so displace-

ments and rotations are constrained to be compatible.

Problem 2 (S46A):

Two groups of PLANE2D, plane stress, 8-node elements are coupled together as
shown in Figure S46–2 where the geometry and material properties are the same as
those in Problem 1. The focus of interest in on the continuum-to-continuum
constraint and the location of the primary point which is no longer located at the
middle of the 3-point curve, but at any arbitrary position.

Problem 3 (S46B):

Two groups of SHELL9 elements are coupled together as shown in Figure S46–3
where the geometry and material properties are the same as those in Problem 1 and
2. The primary deformation is located in the x-y plane. This problem is provided to
verify the accuracy of the structure-to-structure constraint.

ANALYTICAL SOLUTION:

= -pL

/ 3EI

= -pL

/ 2EI

S46, S46A, S46B: Bending of a Cantilever Beam

See

List

= 1 in

= 10 in

= 1/12 in

= lE6 psi

= 0.3

= -1 lb

Chapter 2 Linear Static Analysis

2-72

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

At the free end:

Figure S46-1

inch

rad

Theory

-4.000E-3

-6.000E-4

Beam Element

-4.000E-3

-6.000E-4

Plane Stress Element

-4.006E-3

-6.000E-4

Beam/Plane Stress Element (S46)

-4.008E-3

-6.000E-4

Plane Stress/Plane Stress Element (S46A)

-4.009E-3

-5.985E-4

SHELL9/SHELL9 Element (S46B)

-4.014E-3

-5.990E-4

Computed using displacements at the free end.

E,I

Problem Sketch

Finite Element Model

Figure S46-2

Plane Stress (PLANE2D) Elements

Point-to-Line Constraints

Finite Element Model - 2

Shell (SHELL9) elements

Point-to-line Constraints

Finite Element Model - 3

Figure S46-3

COSMOSM Basic FEA System

2-73

Part 2 Verification Problems

TYPE:

Static analysis, SOLID elements, TETRA10 elements, BEAM elements, and point-
to-surface constraint elements (attachment).

PROBLEM:

Calculate the maximum deflection and the maximum rotation 0 of a cantilever beam
loaded by a shear force at the free end.

GIVEN:

MODELING HINTS: Continuum-to-Structure Constraint

Problem 1 (S47):

The solid elements are defined by nodes 1 through 24. The beam element is defined
by nodes 25 and 26. Each solid element is theoretically equivalent to a beam element
where I = 1/12 in

about y and z axes. Node 25 is attached to surface 21-22-24-23,

so displacements and rotations are constrained to be compatible.

Problem 2 (S47A):

Two groups of SOLID 20-node elements are coupled together as shown in Figure
S47-2 where the geometry and material properties are the same as those in Problem
1. The focus of interest is on the continuum-to-continuum constraint and the location
of the primary point which is no longer located at the middle of the 8-point surface,
but at any arbitrary position.

Problem 3 (S47B):

Two groups of TETRA10 elements are coupled together as shown in Figure S47-3
where the geometry and material properties are the same as those in Problem 1 and
2. This problem is provided to verify the accuracy of the continuum-to-continuum
constraint with the primary point located at any arbitrary position of a 6-node
surface.

ANALYTICAL SOLUTION:

δ = -PL

/ 3EI

θ = -PL

/ 2EI

S47, S47A, S47B: Bending of a Cantilever Beam

See

List

= 1 in

w = 1 in
L

= l0 in

I = 1/12 in

(about y and z axes)

= 1E6 psi

= 0

= 1 lb

Chapter 2 Linear Static Analysis

2-74

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

At the free end.

Figure S47-1

Deflection

inch

Rotation

rad

Theory

-4.000E-3

-6.000E-4

Beam Element

-4.000E-3

-6.000E-4

Plane Stress Element

-4.010E-3

-6.000E-4

Beam/Solid Element (S47)

-4.005E-3

-6.000E-4

Solid/Solid Element (S47A)

-3.986E-3

-5.962E-4 *

Tetra10/Tetra10 Element (S47B)

-3.969E-3

-5.950E-4 *

Computed using displacements at the free end.

L/2

Finite Element Model – 1

L/2

Problem Sketch

Figure S47-2

SOLID Elements

Point-to-surface Constraints

Finite Element Model - 2

Figure S47-3

TETRA10 Elements

Point-to-surface Constraints

Finite Element Model - 3

COSMOSM Basic FEA System

2-75

Part 2 Verification Problems

TYPE:

Static analysis, axisymmetric elements (PLANE2D).

REFERENCE:

Brenkert, Jr., K., “Elementary Theoretical Fluid Mechanics,” John Wiley and Sons,
Inc., New York, l960.

PROBLEM:

A large cylindrical tank is partially filled with an incompressible liquid. The tank
rotates at a constant angular velocity about its vertical axis as shown. Determine the
elevation of the liquid surface relative to the center (lowest) elevation for various
radial positions. Also, determine the pressure p in the fluid near the bottom corner
of the tank.

GIVEN:

COMPARISON OF RESULTS:

S48: Rotation of a Tank of Fluid (PLANE2D Fluid)

See

List

w = l rad/sec
r =

= 20 in

= 0.9345E-4 lb-sec

/in

= 386.4 in/sec

= 30E4 psi

Where:

= bulk modulus

= acceleration due to gravity

ρ = density

Displacements

inch

Pressure (psi)

Node 4

Node 7

Node 11

Element 60

Theory

-1.86335

+1.86335

-0.74248

COSMOSM

-1.8627

+1.8627

-0.74250

Difference

0.036%

0.03%

0.003%

After subtracting from the displacement at Node 1 (-1.4798 in)

Chapter 2 Linear Static Analysis

2-76

COSMOSM Basic FEA System

Figure S48-1

10 11 12 13

Finite Element Model

Problem

Sketch

COSMOSM Basic FEA System

2-77

Part 2 Verification Problems

TYPE:

Static analysis plane strain (PLANE2D) or SOLID elements.

REFERENCE:

Brenkert, K., Jr., “

Elementary

Theoretical Fluid Mechanics,” John Wiley and Sons,

Inc., New York, l980.

PROBLEM:

Large rectangular tank is partially filled with an incompressible liquid. The tank has
a constant acceleration to the right, as shown. Determine the elevation of the liquid
surface relative to the zero acceleration elevation for various Y-axis positions. Also,
determine the slope of the surface and the pressure p in the fluid near the bottom
right corner of the tank.

GIVEN:

COMPARISON OF RESULTS:

S49A, S49B: Acceleration of a Tank

of Fluid (PLANE2D Fluid)

See

List

= 45 in/sec

= 48 in

= 20 in

= 386.4 in/sec

= 30E4 psi

= 0.9345E-4 lb-sec

/in

Where:

= bulk modulus

= acceleration due to gravity

= density

Displacements

inch

Pressure (psi)

Node 3

Node 7

Node 11

Element 60

Theory

-1.86335

+1.86335

0.74248

COSMOSM

-1.8627

+1.8627

0.7425

Difference

0.036%

0.03%

Chapter 2 Linear Static Analysis

2-78

COSMOSM Basic FEA System

Figure S49A-1

Problem Sketch

Finite Element Model

(Use PLANE2D Elements)

9 10 11 12

COSMOSM Basic FEA System

2-79

Part 2 Verification Problems

TYPE:

Static analysis, multi-field elements (4-node PLANE2D, 8-node PLANE2D,
SHELL4T, 6-node TRIANG, 8-node SOLID, 20-node SOLID, TETRA4R and
SHELL6 elements).

REFERENCE:

Roark, R. J., “Formulas for Stress and Strain,” 4th Edition, McGraw-Hill Book Co.,
New York, l965, pp. 166.

PROBLEM:

A curved beam is clamped at one
end and subjected to a shear force
P at the other end. Determine the
deflection at the free end.

GIVEN:

COMPARISON OF RESULTS:

Deflections at free end by theoretical solution is equal to 0.08854 in

S50A, S50B, S50C, S50D, S50F, S50G,

S50H, S50I: Deflection of a Curved Beam

See

List

= 10E6 psi

= 0.25

= 4.12 in

= 4.32 in

= 0.1 in

= 1 lb

Element

COSMOSM

Difference (%)

PLANE2D (4-Node) (S50A)

0.08761

1.05%

PLANE2D (8-Node) (S50B)

0.08850

SHELL4T (S50C)

0.08827

0.26%

TRIANG (6-Node) (S50D)

0.07049

11.6%

TETRA4R (4-Node) (S50H)

0.08785

0.8%

SOLID (8-Node) (S50F)

0.08726

1.45%

SOLID (20-Node) (S50G)

0.08848

0.07%

SHELL6 (Curved) (S50I)

0.07498

15.32%

SHELL6 (Assembled) (S50I)

0.062679

29.2%

Figure S50-3

Problem Sketch

Chapter 2 Linear Static Analysis

2-80

COSMOSM Basic FEA System

TYPE:

Static analysis, beam elements (BEAM2D).

REFERENCE:

Valerian Leontovich, “Frames and Arches,”
McGraw-Hill Book Co., Inc., New York,
l959, pp. 68.

PROBLEM:

Determine the support reactions for frame
shown in the figure.

GIVEN:

= 16 ft

= 8 ft

= 4.32E6 lb/ft

= 6 ft

= 10 lb/ft

= I

= A

Total Load = 4 lbs

MODELING HINTS:

Find load intensity along the frame from

W = (Total load) / q = 4 lb/ft

Then use beam loading commands to solve the problem

COMPARISON OF RESULTS:

Reactions (lb):

S51: Gable Frame with Hinged Supports

See

List

Node

No.

Theory

COSMOSM

4.44

30.00

4.40

30.00

-4.44

10.00

-4.40

10.00

Figure S51-1

L/2

Problem Sketch

I 2

W = q L/2

I 2

Finite Element Model

COSMOSM Basic FEA System

2-81

Part 2 Verification Problems

TYPE:

Static analysis, beam elements (BEAM2D).

REFERENCE:

Morris, C. H. and Wilbur, J. B., “Elementary Structural Analysis,” McGraw-Hill
Book Co., Inc., Second Edition, New York, l960, pp. 93-94.

PROBLEM:

Determine the support reactions for the simply supported beam with intermediate
forces and moments.

Figure S52-1

S52: Support Reactions for a Beam with

Intermediate Forces and Moments

See

List

75K

4K/ft

40K

30K

4K/ft

30K

Problem Sketch

40K 75K

24K ft

Finite Element Sketch

Chapter 2 Linear Static Analysis

2-82

COSMOSM Basic FEA System

GIVEN:

A = 0.3472 ft

= I

= 0.02009 ft

= 0.040l9 ft

= 4320 x l0

K/ft

NOTE:

The sign convention for the intermediate loads follows the local coordinate system
for the beam (defined by the I, J, K nodes).

COMPARISON OF RESULTS:

Figure S52-2

NOTE:

The results obtained with COSMOSM are compared with those given in reference.
The numbers shown in parenthesis are from COSMOSM.

R =

-45K (-45K)

R = 57 K (57K)

R = 89K (89K)

COSMOSM Basic FEA System

2-83

Part 2 Verification Problems

TYPE:

Static analysis, beam elements (BEAM2D).

REFERENCE:

Norris, C. H., and Wilbur, J. B., “Elementary Structural Analysis,” 2nd ed.,
McGraw-Hill Book Co., Inc., l960, pp. 99.

PROBLEM:

Find the reactions in the support and forces and moments in the beam.

Figure S53-1

GIVEN:

= I

= l ft

= 2 ft

A = 3.464 ft

= 432 x l0

k/ft

S53: Beam Analysis with Intermediate Loads

See

List

Finite Element Model

2K/ft

Problem Sketch

Chapter 2 Linear Static Analysis

2-84

COSMOSM Basic FEA System

NOTE:

The sign convention for intermediate loads, follows the local coordinate system, for
the beam (defined by I, J, K nodes).

COMPARISON OF RESULTS:

Figure S53-2

NOTE:

COSMOSM results are given in parentheses.

33.33 K.ft.

11.13 K
(11.13 K)

20.87 K
(20.87 K)

10.0 K
(10.0 K)

0.0 K
(0.0 K)

(-33.33 K.ft.)

(33.33 K.ft)

Node 1

Nod 2

Node 2

Node 3

Element 1

Element 2

Use the BEAMRESLIST (Results, List, Beam End Force)
command to list the results

COSMOSM Basic FEA System

2-85

Part 2 Verification Problems

TYPE:

Static analysis, beam elements (BEAM2D).

REFERENCE:

Weaver, Jr., W., and Gere, J. M., “Matrix Analysis of Framed Structures,” 2nd ed.,
D. Van Nostrand Company, New York, l980. pp. 280, 486.

PROBLEM:

Find the deformations and forces in the plane frame subjected to intermediate forces
and moments.

Figure S54-1

GIVEN:

= 0.04 m

= I

= 2 x 10

= 4 x 10

= 200 x 10

KN/m

= 3m

= 30 x 2

(1/2)

= 60 KN

M = 180 KN-m

S54: Analysis of a Plane Frame with Beam Loads

See

List

L/2

Problem Sketch

Finite Element Model

40.75*

40.75

4.747

5.44

* Results obtained from reference

+ Results obtained from COSMOS/M

Moments are in KNm units

Figure S54-2

Chapter 2 Linear Static Analysis

2-86

COSMOSM Basic FEA System

TYPE:

Static analysis, beam element (BEAM3D).

REFERENCE:

Crandall, S. H., and Dahl, N. C., “An Introduction to the Mechanics of Solids,”
McGraw-Hill Book Co., Inc., New York, l959, pp. 342.

PROBLEM:

A cantilever beam of width b and length L has a depth which tapers uniformly form
d at the tip to 3d at the wall. It is loaded by a force P at the tip. Find the maximum
bending stress at x = L (midspan).

Figure S55-1

S55: Laterally Loaded Tapered Beam

See

List

GIVEN:

= 4000 lb

= 50 in

= 3 in

= 2 in

= 30E6 psi

COMPARISON OF RESULTS:

(psi)

(at node 2)

Theory

8333

COSMOSM

8333

L/2

Problem Sketch

Finite Element Model

COSMOSM Basic FEA System

2-87

Part 2 Verification Problems

TYPE:

Static analysis, 9-node shell element (SHELL9).

REFERENCE:

Hughes, T. J. R., Taylor, R. L., and Kanoknukulchai, W. A., “Simple and Efficient
Finite Element for Plate Bending,” I.J.N.M.E., 11, 1529-1543, 1977.

PROBLEM:

A circular thick plate clamped at the boundary is subjected to a point load at its
center. (Shown in Figure S56-1).

Determine the transverse displacement along the radius r.

GIVEN:

= 1.09E6

ν = 0.3
t

= 2 (thickness) in

= 4 lb

= 5 in

ANALYTICAL SOLUTION:

Where:

D = Et

/ 12(1-

)

G = E / 2 (1+

ν)

K = 0.8333 (shear correction factor)

S56: Circular Plate Under a Concentrated

Load (SHELL9 Element)

See

List

Chapter 2 Linear Static Analysis

2-88

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

Figure S56-1

Node

r (in)

max

(in) x 10

-6

Analytical

COSMOSM

0.0

—

-6.469

.625

4.185

-4.242

1.250

3.1670

-3.166

1.875

2.3474

-2.349

2.500

1.6366

-1.637

3.125

1.0316

-1.033

3.750

0.5458

-0.5465

4.375

0.1962

-0.1968

Finite Element Model

(12 Elements)

Problem Sketch

COSMOSM Basic FEA System

2-89

Part 2 Verification Problems

TYPE:

Static analysis, 9-node shell element (SHELL9).

REFERENCE:

Dvorkin, E. N., and Bathe, K. J., “A Continue Mechanics Based Four Node Shell
Element for General Nonlinear Analysis,” Engineering Computations, 1, 77-78,
1984.

PROBLEM:

A cylindrical shell with both ends covered with rigid diaphragms which allow
displacement only in the axial direction of the cylinder is subjected to a concentrated
load on the center (shown in the figure below). Determine the radial deflection of
point P.

MODELING HINTS:

Due to symmetry, only one-eighth of the cylinder is modeled. To simulate the rigid
diaphragm, on the boundary of the cylinder with z = 0, no rotation along the axial
direction (z-axis) is allowed.

S57: Test of a Pinched Cylinder with

Diaphragm (SHELL9 Element)

See

List

GIVEN:

= 300 in

= 600 in

= 3E6 psi

= 0.3

= 3 (thickness) in

= 1 lb

COMPARISON OF RESULTS

(inch)

Theory

0.18248E-4

COSMOSM

0.17651E-4

Chapter 2 Linear Static Analysis

2-90

COSMOSM Basic FEA System

Figure S57-1

Finite Element Model

(4 x 4)

P/4

Diaphragm

Problem Sketch

L/2

COSMOSM Basic FEA System

2-91

Part 2 Verification Problems

TYPE:

Static analysis, 9-node shell element (SHELL9), 4-node tetrahedral element
(TETRA4R).

REFERENCE:

MacNeal, R. H. and Harder, R. L., “A Proposed Standard Set of Problems to Test
Finite Element Accuracy,” F. E. in Analysis and Design, pp. 3-20, 1986.

PROBLEM:

A twisted beam is subjected to a concentrated load at the tip in the in-plane and out-
of-plane directions (shown in the figure below). Determine the deflections
coincident with the load.

GIVEN:

= 12 in

W = 1.1 in
h

= 0.32, 0.0032 (thickness) in

= 1 lb for h = 0.32 and 1e-6 lb for h=0.0032

= 29E6 psi

= 0.22

COMPARISON OF RESULTS:

S58A, S58B, S58C: Deflection of a Twisted Beam

with Tip Force

See

List

Deflection in the direction of the

force

Thickness (in)

Force Direction

Theory

COSMOSM

h = 0.32 in

(S58A: SHELL9)

Force = 1.0 lb

In-Plane (Load case 1)

Out-of-Plane (Load Case 2)

0.5240E-2
0.1754E-2

0.5397E-2
0.1759E-2

h = 0.0032 in

(S58B: SHELL9)

Force=1e-6 lb

In-Plane (Load case 1)

Out-of-Plane (Load Case 2)

0.5256E-2
0.1794E-2

0.4704E-2
0.1255E-2

h = 0.32

(S58C: TETRA4R)

Force = 1.0 lb.

In-Plane (Load case 1)

Out-of-Plane (Load Case 2)

0.5240E-2
0.1754E-2

0.4967E-2
0.1600E-2

Chapter 2 Linear Static Analysis

2-92

COSMOSM Basic FEA System

Figure S58-1

F (out-of-plane)

Twist = 90

Problem Description and

Finite Element Mesh (1 x 6)

F (in-plane)

COSMOSM Basic FEA System

2-93

Part 2 Verification Problems

TYPE:

Static analysis, 4- and 9-node composite shell elements (SHELL4L, SHELL9L),
solid composite element (SOLIDL).

REFERENCE:

Chang, T. Y., and Sawamiphakdi, K., “Large Deformation Analysis of Laminated
Shells by Finite Element Method,” Computers and Structures, Vol. 13, pp. 331-340,
1981.

PROBLEM:

A square sandwich plate consisting of two identical facings and an aluminum
honeycomb core is subjected to uniform loading as shown in the figure below.
Determine the central deflection of the plate at point A.

MODELING HINTS:

Due to symmetry, one quarter of the plate is modeled. To ensure computational
stability, a small elastic modulus (E = 1.0E-12) for the core is used.

S59A, S59B, S59C: Sandwich Square Plate

Under Uniform Loading (SHELL9L)

See

List

GIVEN:

Facing:
E

= 10.5E6 ksi

= 0.3

hf = 0.015 in (thickness)
Core:
E

= 0 ksi

= 25 in

= G

= 50 ksi

= 9.2311 psi

hc = 1 in (thickness)

COMPARISON OF RESULTS:

max

at the Center

Reference
SHELL4L (S59B)
SHELL9L (S59A)
SOLIDL (S59C)

0.846
0.868
0.849

COSMOSM
SHELL4L (S59B)
SHELL9L (S59A)
SOLIDL (S59C)

0.851
0.866
0.849

Chapter 2 Linear Static Analysis

2-94

COSMOSM Basic FEA System

Figure S59A-1

clamped

Problem Sketch and Finite Element Model

Cross Section

2a = 50

COSMOSM Basic FEA System

2-95

Part 2 Verification Problems

TYPE:

Static analysis, 9-node shell element (SHELL9).

REFERENCE:

Timoshenko, S. P. and Woinowsky-Krieger, S., “Theory of Plates and Shells,” 2nd
Ed., McGraw Hill, New York, 1959.

PROBLEM:

Determine the maximum deflection (at point A) of a clamped-clamped plate (shown
in the figure below) with uniform loading and modeled by a skewed mesh. Various
span-to-depth ratios are investigated.

GIVEN:

= 1E7 psi

= 0.3

= 2 in

= 1 psi (0.01 psi is used for thickness 0.002)

= thickness = 0.2, 0.02, and 0.002 in

MODELING HINTS:

Due to symmetry, only one quarter of the plate is modeled.

ANALYTICAL SOLUTION:

= 0.00126 qa

Where:

D = Et

/ 12(1 -

)

S60: Clamped Square Plate Under

Uniform Loading

See

List

Chapter 2 Linear Static Analysis

2-96

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

Figure S60-1

Span/Thickness Ratio

Deflection (inch)

Theory

COSMOSM

(q = 1.0 psi)

-2.7518E-6

-3.4758E-6

100

(q = 1.0 psi)

-2.7518E-3

-3.0649E-3

1,000

(q = 0.01 psi)

-2.7518E-2

-2.79259E-2

The input file provided (S60.GEO) is for a span/thick-

ness ratio of 10. You need to redefine the thickness for
other ratios using the RCONST command.

Better accuracy can be obtained with a finer mesh.

Finite Element Model

(2 x 2 Skew)

Problem Sketch

COSMOSM Basic FEA System

2-97

Part 2 Verification Problems

TYPE:

Static analysis, crack element, stress intensity factor, 8-node plane continuum
element (PLANE2D).

REFERENCE:

Brown, W. F.,

Jr., and Srawley, J. E., “Plane Strain Crack Toughness Testing of High

Strength Metallic Materials,” ASTM Special Technical Publication 410,
Philadelphia, PA, 1966.

PROBLEM:

Determine the stress
intensity factor of a
single-edge-cracked
bend specimen using
the crack element.

GIVEN:

= 30 x 10

psi

= 0.3

Thickness = 1 in
a

= 2 in

= 4 in

= 32 in

= 1 lb

COMPARISON OF RESULTS:

S61: Single-Edge Cracked Bend Specimen,

Evaluation of Stress Intensity Factor

Using Crack Element

See

List

Theory

10.663

COSMOSM

9.855

Problem Sketch

Finite Element Model

P/2

Figure S61-1

Chapter 2 Linear Static Analysis

2-98

COSMOSM Basic FEA System

TYPE:

Static analysis, crack stress intensity factor, 8-node plane continuum element
(PLANE2D).

REFERENCE:

Cook, Robert D., and Cartwright, D. J., “Compendium of Stress Intensity Factors,”
Her Majesty’s Stationary Office, London, 1976.

PROBLEM:

Determine the
stress intensity
factor of the center-
cracked plate.

GIVEN:

= 30x 10

psi

= 0.3

Thickness = 1 in
W = 20 in
a

= 2 in

= 1 lb/in

COMPARISON OF RESULTS:

S62: Plate with Central Crack

See

List

Theory

2.5703

COSMOSM

2.668

Problem Sketch

Finite Element Model

Figure S62-1

COSMOSM Basic FEA System

2-99

Part 2 Verification Problems

TYPE:

Static analysis, cyclic symmetry, truss elements (TRUSS2D).

REFERENCE:

Cook, Robert D., “Concepts and Applications of Finite Element Analysis.” 2nd
Edition, John Wiley & Sons, New York, 1981.

PROBLEM:

The pin-jointed plane
hexagon is loaded by
equal forces P, each
radial from center 0.
All lines are uniform
and identical. Find the
radial displacement of
a typical node.

GIVEN:

= 120 in

= l20 in

A = l0 in

= 3000 lb

= 30E6 psi

MODELING HINTS:

Taking advantage of the cyclic symmetry of the model and noting that the model
displaces radically the same amount at all six nodes, only one element is considered
with the radial degree of freedom coupled in the cylindrical coordinate system.

COMPARISON OF RESULTS:

Radial Displacement = 2PL/AE = (3000)(120)/(10)(30E6) = 0.0012 in

S63: Cyclic Symmetry Analysis

of a Hexagonal Frame

See

List

Radial Displacement

Theory

0.0012 in

COSMOSM

0.0012 in

Geometric Model

Finite Element Model

Figure S63-1

Chapter 2 Linear Static Analysis

2-100

COSMOSM Basic FEA System

TYPE:

Static analysis, cyclic symmetry, 3-node triangular elements (TRIANG).

REFERENCE:

Cook, Robert D., “Concepts and Applications of Finite Element Analysis.” 2nd
Edition, John Wiley & Sons, New York, 1981.

PROBLEM:

A hexagonal shaped plate is loaded by a set of
radial forces as shown in the figure below.
Calculate the deformation of the structure at the
point where the load is applied. The plate is
considered as a plane stress problem and
modeled with 3-node triangular plane elements.

GIVEN:

= l0 in

= l in

= 3000 lb

= 30E6 psi

MODELING HINTS:

This plate is built by combining six sub-structures
at 60 degree angles relative to one another. Taking
advantage of the cyclic sub-structure may be considered for analysis. Note that for
the sub-structure shown, the displacements of nodes along A-A and B-B must be the
same in the radial directions. Therefore, these nodes will be coupled radially in the
cylindrical coordinate system. All degrees of freedom in the circumferential
direction will be fixed.

COMPARISON OF RESULTS:

The problem is solved for both the full structure and the sub-structure with the
displacements coming out identical for the corresponding nodes.

S64A, S64B: Cyclic Symmetry

See

List

Displacement at Point A

X-Displacement

Y-Displacement

Full Model (S64A)

-9.445E-5

-1.636E-4

Cyclic Part (S64B)

-9.450E-5

-1.637E-4

Full Structure

Sub-Structure

Figure S64-1

COSMOSM Basic FEA System

2-101

Part 2 Verification Problems

TYPE:

Static analysis, axisymmetric solid (PLANE2D) and fluid (PLANE2D) elements.

REFERENCE:

S. Timoshenko and S. Woinowsky-Kreiger, “Theory of Plates and Shells,” 2nd
Edition, McGraw Hill, New York, 1959, pp. 485-487.

PROBLEM:

A large cylindrical tank is filled with an incompressible liquid. The tank rotates at a
constant velocity about its vertical axis as shown. Determine the deflection of the
tank wall and the bending and shear stresses at the bottom of the tank wall.

S65: Fluid-Structure Interaction,

Rotation of a Tank of Fluid

See

List

GIVEN:

= 48 in

= 20 in

= 1 in

Fluid:
ρ

= 0.9345E-4 lb-sec

/in

= 30E4 psi

Tank:
E

= 3E7 psi

= 0.3

ω = 1 rad/sec
g

= 386.4 in/sec

Where:
b

= Bulk modulus

= Accel. due to gravity

= Density

= Young's modulus

= Poisson’s ratio

COMPARISON OF RESULTS:

Deflection in x-direction (10

-6

in)

Y (in)

Point

Theory

COSMOSM

20
16
12

8
4
0

A
B
C
D
E
F

7.381
19.544
27.805
27.161
13.604

8.269
18.854
26.870
26.578
13.700

Theory

COSMOSM

yy,

(max), psi

52.24

40.34

, lb/in

3.76

3.20

Note: Compatibility is imposed along the
direction normal to the interface using the
CPDOF command (LoadsBC, Structural,
Coupling, Define DOF Set).

Chapter 2 Linear Static Analysis

2-102

COSMOSM Basic FEA System

ANALYTICAL SOLUTIONS:

Deflection w(y):

λ = ρg
P = pressure applied on the tank wall due to an angular velocity

End Moment M

End Shear force Q

Figure S65-1

Liquid

Problem Sketch

Interface

48 Elements

4 Elements

20
Elements

Finite Element Model

COSMOSM Basic FEA System

2-103

Part 2 Verification Problems

TYPE:

Static analysis, plane strain solid (PLANE2D) and plane fluid (PLANE2D)
elements.

REFERENCE:

Timoshenko, S. P., and Gere, James M., “Mechanics of Materials,” McGraw Hill,
New York, 1971, pp. 167-211.

PROBLEM:

A large rectangular tank is filled with an incompressible liquid. The tank has a
constant acceleration to the right, as shown. Determine the deflection of the tank
walls and the bending and shear stresses at the bottom of the right tank wall.

S66: Fluid-Structure Interaction,

Acceleration of a Tank of Fluid

See

List

GIVEN:

r = 48 in
h = 20 in
t = 1 in

Fluid:
ρ =

0.9345E-4

lb-sec

/in

b = 30E4 psi

Tank:
E= 3E7 psi
ν = 0.3
a = 45 in/sec

g = 386.4 in/sec

Where:
b = Bulk modulus
g = Gravity Accel.
ρ = Density
E= Young's modulus
ν = Poisson’s ratio

COMPARISON OF RESULTS:

Point Y (in)

(inch)

Theory

COSMOSM

Theory

COSMOSM

A
B
C
D
E
F

20
16
12

8
4
0

2.137E-3
1.591E-3
1.054E-3
5.564E-4
1.662E-4

2.150E-3
1.602E-3
1.062E-3
5.619E-4
1.689E-5

6.667E-4
5.120E-4
3.552E-4
1.989E-4
6.348E-4

6.761E-4
5.193E-4
3.605E-4
2.024E-4
6.512E-5

Theory

COSMOSM

, (max), psi (at y = 0)

410.02

386.39

, lb/in (at y = 0)

9.24

8.84*

*This value is calculated by averaging TXY at the nodes
located at the bottom of the right wall giving half weight
to corner nodes.

Note: Compatibility is imposed along the direction
normal to the interface using the CPDOF command
(LoadsBC, Structural, Coupling, Define DOF Set).

Chapter 2 Linear Static Analysis

2-104

COSMOSM Basic FEA System

ANALYTICAL SOLUTIONS:

Deflections of the right wall W

(y) and the left wall W

(y):

Where:

= pgh

= pressure applied on the right wall due to acceleration

= pressure applied on the left wall due to acceleration

= E/(1-

)

End Moment M

= -EI d

W / dy

End Shear Force V

τA

= -EI d

W / dy

Figure S66-1

Liquid

Problem Sketch

a = 45 in/sec

Interface

Elements

Finite Element Model

Elements

1235

1135

105

101

20
Elements

1139

1239

COSMOSM Basic FEA System

2-105

Part 2 Verification Problems

TYPE:

Static analysis, plane stress quadrilateral p-element (8-node PLANE2D) with the
polynomial order of shape functions equal to 5.

PROBLEM:

Calculate the maximum deflection of a cantilever beam loaded by a concentrated
end force.

GIVEN:

COMPARISON OF RESULTS:

Figure S67-1

S67: MacNeal-Harder Test

See

List

Geometric Properties:

= 0.2 in

= 0.1 in

= 6 in

= 2/3 x 10

-4

Material Properties:

= 1 x 10

psi

= 0.3

Loading:

= 1 lb

Theory

COSMOSM

Tip Displacement

0.1081 in

0.10807 in

Chapter 2 Linear Static Analysis

2-106

COSMOSM Basic FEA System

TYPE:

Static analysis, plane stress quadrilateral (8-node PLANE2D) and triangular (6-node
TRIANG) p-elements with the polynomial order of shape functions equal to 5.

PROBLEM:

Calculate the maximum stress of a plate with a circular hole under a uniformly
distributed tension load.

Material Properties:

= 30 x 10

psi

= 0.3

Loading:

= 1000 psi

Figure S68-1

S68: P-Method Solution of a Square

Plate with Hole

See

List

GIVEN:

COMPARISON OF RESULTS:

Geometric Properties:

= 12 in

= 1 in

Theory

COSMOSM
(PORD = 5)

Max. Stress in

X-Direction

3018

3058 psi

•

COSMOSM Basic FEA System

2-107

Part 2 Verification Problems

TYPE:

Static analysis, plane stress triangular p-element (6-node TRIANG).

REFERENCE:

Barlow, J., and Davis, G. A. O., “Selected FE Benchmarks in Structural and Thermal
Analysis,” NAFEMS Rept. FEBSTA, Rev. 1, October, 1986, Test No. LG1.

PROBLEM:

Calculate the stresses at point D of an elliptic membrane under a uniform outward
pressure.

Figure S69-1

S69: P-Method Analysis of an Elliptic

Membrane Under Pressure

See

List

GIVEN:

COMPARISON OF RESULTS

E = 210 x 10

MPa

ν = 0.3
t = 0.1
p = 10 MPa

, at Point D

Theory

92.70

COSMOSM

93.72

1.0

1.75

2.0

1.25

All dimensions in meters

Thickness = 0.1

3.25

2.75

= 1

+ = 1

(

)

(

)

(

)

Chapter 2 Linear Static Analysis

2-108

COSMOSM Basic FEA System

TYPE:

Linear thermal stress analysis, plane continuum element (PLANE2D).

PROBLEM:

A flat plate consists of different material properties through its length. Determine the
deflections and thermal stresses in the plate due to uniform changes of temperature
equal to 100

° F and 200° F.

Figure S70-1

S70: Thermal Analysis with Temperature

Dependent Material

See

List

GIVEN:

t =

0.1

= 0.00001 in/in/

°F

= 0

= 30,000 ksi

COMPARISON OF RESULTS:

for All Elements

T = 100

F *

T = 200

Theory

-30 ksi

-48 ksi

COSMOSM

-30 ksi

-48 ksi

* The temperature in the input file corresponds to T =

200

F. You need to delete the applied temperature

using the NTNDEL command and apply tempera-
ture of 100

F using the NTND command.

(ksi)

30000

20000

150

200

Temperature

Elements 3 & 4

Elements 1 & 2

E = 30E3 ksi (CONSTANT)

Finite Element Model

COSMOSM Basic FEA System

2-109

Part 2 Verification Problems

TYPE:

Static analysis, composite shell element (SHELL4L).

PROBLEM:

Determine the total deflection of the sandwich beam subjected to a concentrated
load.

GIVEN:

THEORY:

D = E

/6 + E

btd

/2 + E

/12 = 8.28 X 10

N.mm

δ = WL

/48D + WLc/4bd

G = 6.2902 + 3.986 = 10.276 mm

COMPARISON OF RESULTS:

Figure S71-1

S71: Sandwich Beam with Concentrated Load

See

List

= 7000 N/mm

= 3 mm

= 1000 mm

= 20 N/mm

= 25 mm

= 100 mm

= 5 N/mm

= 28 mm

W = 250 N

Midspan

Deflection (mm)

Theory

10.276

COSMOSM

10.323

Chapter 2 Linear Static Analysis

2-110

COSMOSM Basic FEA System

TYPE:

Static analysis, tetrahedral elements (TETRA4, TETRA4R).

PROBLEM:

Constraint displacements at one end and prescribed displacements at the other end
of the plate to produce a constant stress state with

= 0.1667E5 and

= 0.

Patch test model.

GIVEN:

Ex = 1E8
υ

= 0.25

δx = 0.4E-2
t

= 0.024

= 0.12

b =

0.24

RESULTS:

All the above elements pass
the patch test. The nodal
stresses show that

0.1667E5 and

= 0.

S74: Constant Stress Patch Test (TETRA4R)

See

List

Figure S74-1

Finite Element Model for Patch Test

COSMOSM Basic FEA System

2-111

Part 2 Verification Problems

TYPE:

Linear static analysis, beam and gap elements (BEAM2D, GAP).

PROBLEM:

The problem is modeled using BEAM2D elements. Five gap elements with zero gap
distances are used. Two different load cases were selected, and the analysis was
performed.

GIVEN:

beam

= 30 x 10

psi

= 1.2 in

= 10 in

= 100 in

= 50 in

COMPARISONS OF RESULTS:

The deformation state of gaps for each load case agrees with the beam deformed
shape corresponding to that load case. The results can be compared with the solution
obtained from linear static analysis, where the gaps are removed and the nodes at the
closed gaps are fixed.

OBTAINED RESULTS:

S75: Analysis of a Cantilever Beam with Gaps,

Subject to Different Loading Conditions

See

List

Forces in Gap Elements

Applied Forces

Load Case

Gap 1

Gap 2

Gap 3

Gap 4

Gap 5

Fa = -1000
Fc = -2000

-361.84

-1197.4

-842.11

Fa = -1000
Fb = -1000

-1206.3

-275.0

Chapter 2 Linear Static Analysis

2-112

COSMOSM Basic FEA System

Figure S75-1

Problem Sketch

FInite Element Model

Load Case 1

Fa = - 1000
Fc = - 2000

Load Case 2

Fa = - 1000
Fb = - 1000

COSMOSM Basic FEA System

2-113

Part 2 Verification Problems

TYPE:

Linear static analysis, beam, plane and gap elements (BEAM2D, PLANE2D,
TRUSS2D and GAP).

PROBLEM:

The shape of the piston is simulated through gap distances. In order to avoid
singularities in the structure stiffness, two soft truss elements are used to hold the
piston. The problem is analyzed for two different pressure values.

GIVEN:

Gap Distances:

= g

= 0.027 in

= g

= 0.001 in

= g

= 0.008 in

= 0 in

= 10 in

= 1.2 in

k = 1 lb/in
E = 30 x 10

psi

Load case 1:

P = 52.5 psi

Load case 2:

P = 90.8 psi

COMPARISON OF RESULTS:

The forces in the gap elementss at a particular of time are in good agreement with
the total force applied to the piston at that time. The deformed shape of the beam for
each load case is comaptible with the forces and location of closed gaps for that load
case.\

S76: Simply Supported Beam Subject to Pressure

from a Rigid Parabolic Shaped Piston

See

List

Chapter 2 Linear Static Analysis

2-114

COSMOSM Basic FEA System

Figure S76-1

Forces in Gap Elements (lb)

Total Force

(Theory)

No. of Closed Gaps

Pressure

Gap Forces

Total

p = 52.5

-215.5
-215.5
-1668.0
-1668.0

-3767

-3780

p = 90.8

-3259.0
-3259.0

-6518

-6540

120 in

60 in

y = (x/100) 3

Problem Sketch

Finite Element Model

COSMOSM Basic FEA System

2-115

Part 2 Verification Problems

TYPE:

Static analysis, direct material property input, hexahedral solid element (SOLID).

REFERENCE:

Roark, R.J. “Formulas for Stress and Strain”, 4th Edition, McGraw-Hill Book Co.,
New York, 1965, pp. 104-106.

PROBLEM:

A beam of length L, width b, and height h is built-in at one end and loaded at the free
end with a shear force F. Determine the deflection at the free end.

GIVEN:

= 10 in

= 30E6 psi

= 1 in

υ =

0.3

= 2 in

= 300 lb

MODELING HINTS:

Instead of specifying the elastic material properties by E and

υ, the elastic matrix [D]

shown below is provided by direct input of its non-zero terms.

S77: Bending of a Solid Beam Using

Direct Material Matrix Input

See

List

[D]

MC11 MC12 MC13 MC14 MC15 MC16

MC22 MC23 MC24 MC25 MC26

MC33 MC34 MC35 MC36

MC44 MC45 MC46

MC55 MC56

Sym.

MC66

Chapter 2 Linear Static Analysis

2-116

COSMOSM Basic FEA System

RESULTS:

Displacement in Z-direction at the tip using E and

υ, is compared with those

obtained with direct input of elastic coefficients in matrix [D].

Figure S77-1

Using E and

Using Direct

Matrix Input

Theory

0.005

COSMOSM

0.00496

Finite Element

Model

Problem

Geometry

COSMOSM Basic FEA System

2-117

Part 2 Verification Problems

TYPE:

P–adaptive analysis, plane stress triangular p-element (TRIANG).

PROBLEM:

Calculate the maximum stress of a plate with a circular hole under a uniform
distributed tension load.

GIVEN:

Geometric Properties:

= 200 in

= 20 in

= 30 x 10

psi

= 1 in

Loading:

= 1 psi

RESULTS:

Nodes: 33, elements: 12, allowable local displacement error: 5%.

S78: P-Adaptive Analysis of a Square

Plate with a Circular Hole

See

List

Iter.

No.

Min

Max.

d.o.f.

Energy

x 10

-4

Max.

Displ.

x 10

-6

Max.

Stress

Local

Displ.

Error

No. of

Sides Not

Converged

1
2
3
4

Ref.

1
2
2
2
4

1
2
3
4
4

16
56
85

100
133

1.692
1.701
1.704
1.704
1.706

3.468
3.480
3.489
3.490
3.502

1.586
2.418
2.692
2.817
2.994

29.544
12.844

1.083

22
16

8
0

Chapter 2 Linear Static Analysis

2-118

COSMOSM Basic FEA System

Figure S78-1

Figure S78-2

Plate with a Hole

Poly nomial O rder for Each Side at Iteration No. 4

COSMOSM Basic FEA System

2-119

Part 2 Verification Problems

TYPE:

Static linear analysis using the asymmetric loading option (SHELLAX).

REFERENCE:

Zienkiewicz, O. C., “The Finite Element Method,” 3rd Edition, McGraw Hill Book
Co., p 362.

PROBLEM:

Determine the radial displacement
of a hemispherical shell under a
uniform unit moment around the
free edge.

GIVEN:

= 100 in

r =

= 2 in

= 1E7 psi

M = l in-lb/in
ν (NUXY) = 0.33

MODELING HINTS:

It is important to note that nodal load is to be specified per unit radian which in this
case is 50 in-lb/rad.

= Mx Arcx = Mx Rx

Ψ = 1(50) (1) rad = 50]

Where:

Ψ = horizontal angle

COMPARISON OF RESULTS:

S79: Hemispherical Shell Under Unit

Moment Around Free Edge

See

List

Radial Displacement at Node 31

Theory

1.58E-5 in

COSMOSM

1.589E-5 in

Figure S79-1

Problem Sketch

Chapter 2 Linear Static Analysis

2-120

COSMOSM Basic FEA System

TYPE:

Static linear analysis using the asymmetric loading option in SHELLAX.

REFERENCE:

NAFEMS, BranchMark Magazine, November, 1988.

PROBLEM:

Determine the stress of an
axisymmetric Hyberbolic
shell under loading F = cos2

on the outward edge, y = 1.

GIVEN:

= 1 m

tan

φ = 2

(-1/2)

= 2

(1/2)

E =

210E3

MPa

ν (NUXY) = 0.3
Thickness = 0.01

MODELING HINTS:

Due to symmetry only half of the shell will be modeled. The Cosine load at the free
edge will be applied in terms of its x- and y- components, representing the second
term of the even function for a Fourier expansion.

COMPARISON OF RESULTS:

The results in the following table correspond to the NXZ component of stress for
element 1 as recorded in the output file.

S80: Axisymmetric Hyperbolic Shell Under a

Cosine Harmonic Loading on the Free Edge

See

List

Shear Stress (y = 0,

= 45

)

Theory

-81.65 MPa

COSMOSM

-79.63 MPa

Y = r -1

Problem Sketch

Finite Element Model

Figure S80-1

COSMOSM Basic FEA System

2-121

Part 2 Verification Problems

TYPE:

Static linear analysis using the asymmetric loading option in SHELLAX.

REFERENCE:

SHELL4 elements are used for comparison purposes.

PROBLEM:

A circular plate with inner and outer radii of 3 in and 10 in respectively, is subjected
to a non-axisymmetric load around outer circumference from

θ = -54° to θ = 54°

perpendicular to the plate surface. The load distribution is:

F (

θ) = 5.31 [1 + cos(10

/3)]*10

GIVEN:

= 3 in

= 10 in

= 3E7 psi

= 1 in

ν (NUXY) = 0.3

MODELING HINTS:

A total of seven elements are considered in this example. Note that since the load is
symmetric about the x-axis, it will be considered only between

= 0

° and

= 54°

at 3

° intervals, and represented by the even (Cosine) terms of the Fourier expansion.

Only the first six (Cosine) terms will be included.

COMPARISON OF RESULTS:

S81: Circular Plate Under

Non-Axisymmetric Load

See

List

Displacement of Outer Edges

(

= 180

) in the Axial Direction

SHELLAX

5.80 x 10

-4

SHELL4

5.62 x 10

-4

Chapter 2 Linear Static Analysis

2-122

COSMOSM Basic FEA System

Figure S81-1

Problem Sketch

= -54

= +54

Finite Element Model

1 2 3

4 5 6

Load
Distribution

COSMOSM Basic FEA System

2-123

Part 2 Verification Problems

TYPE:

Static analysis, PLANE2D element using asymmetric loading option.

REFERENCE:

Timoshenko, S., “Strength of Materials,” 3rd Edition, D. Van Nostrand Co., Inc.,
New York, 1956.

PROBLEM:

A long solid circular shaft is built-in at one end and subjected to a twisting moment
at the other end. Determine the maximum shear stress,

max, at the wall due to the

moment.

Figure S82-1

S82: Twisting of a Long Solid Shaft

See

List

d/2

_ 0 0

0 0

Finite Element Model

Problem Sketch

Chapter 2 Linear Static Analysis

2-124

COSMOSM Basic FEA System

GIVEN:

= 30E6 psi

= 24 in

= 1 in

M = -200 in-lb

MODELING HINTS:

Since the geometry is axisymmetric about the y-axis,
the finite element model, shown in the figure above, is
considered for analysis. The effect of the applied
moment is calculated in terms of a tangential force
integrated around the circumference of the circular rod.

ANALYTICAL SOLUTION:

The load is applied at (node 63) in the z-direction (circumferential). Ux (radial)
constraints are not imposed at the wall in order to allow freedom of cross-sectional
deformation which corresponds to the assumptions of “negligible shear” stated in
the reference.

COMPARISON OF RESULTS:

At clamped edge (node 3).

Max Shear Stress (psi)

Theory

1018.4

COSMOSM

1018.4

Figure S82-2

COSMOSM Basic FEA System

2-125

Part 2 Verification Problems

TYPE:

Static analysis, PLANE2D element using the asymmetric loading option.

REFERENCE:

Timoshenko, S., “Strength of Materials,” 3rd Edition, D. Van Nostrand Co., Inc.,
New York, 1956.

PROBLEM:

A long solid circular shaft is built-in at one end and at the other end a vertical force
is applied. Determine the maximum axial stress

y at the wall and at one inch from

the wall due to the force.

Figure S83-1

S83: Bending of a Long Solid Shaft

See

List

_ 0 0

0 0

Finite Element Model

d/2

Problem Sketch

Chapter 2 Linear Static Analysis

2-126

COSMOSM Basic FEA System

GIVEN:

= 30E6 psi

= 24 in

= 1 in

= -25 lb

MODELING HINTS:

The finite element model is formed as noted in the figure
considering the axisymmetric nature of the problem. The
force applied at node 63 is calculated based on a Fourier
Sine expansion representing its antisymmetric nature.

ANALYTICAL SOLUTION:

The load is applied at (node 75) in the z-direction (circumferential). Ux (radial)
constraints are not imposed at the wall in order to allow freedom of cross-sectional
deformation which corresponds to the assumptions of “negligible shear” stated in
the reference.

COMPARISON OF RESULTS:

At element 1 and

θ = 90°.

Max Axial Stress (sy psi)

y = 0 (Node 3)

y = 1 in (Node 5)

Theory

6111.6

5856.9

COSMOSM

6115.1

5856.8

Load Distribution

Figure S83-2

COSMOSM Basic FEA System

2-127

Part 2 Verification Problems

TYPE:

Static analysis TRIANG element using the submodeling option.

PROBLEM:

Calculate the maximum von Mises stress for a square plate under a concentrated
load at one corner. Compare the displacement and stress results from a fine mesh to
the results from an originally coarse mesh improved using submodeling.

GIVEN:

= 25 in

= 30 E6 psi

= 25 in

= 0.1 in

Fx = Fy = 1000 lbs

Figure S84-1

COMPARISON OF RESULTS:

S84: Submodeling of a Plate

See

List

Mesh Type

Max Deflection at Node 1

Max Stress

Coarse Mesh

-0.00131

3810

Coarse Mesh + Submodeling

-0.00156

7626

Fine Mesh and Theory

-0.00156

7620

submodel

FX=1000

FY=1000

FIXED

EDGE

FIXED

EDGE

FREE ED

Chapter 2 Linear Static Analysis

2-128

COSMOSM Basic FEA System

TYPE:

Static analysis, SHELL4 plate elements on elastic foundation.

PROBLEM:

A simply supported plate is subjected to uniform pressure P. The full plate is
supported by elastic foundation. For small flexural rigidity, the calculated pressure
applied to the plate from the foundation approaches the applied external pressures.
The flexural rigidity decreases by decreasing the thickness and modulus of elasticity.

Figure S85-1

NOTE:

Foundation pressure is recorded in the output file for each element in the last column
of element stress results.

S85: Plate on Elastic Foundation

See

List

GIVEN:

= 30 x 10 psi

= 0.3

= 0.01 in

= 10 in

= 10 psi

COMPARISON OF RESULTS:

Foundation Pressure

at Element 200

Theory

-10.0

COSMOSM

-10.0

Pressure P

SIMP

LY SU

PPOR

TED

SIMPLY SU

PPOR

TED

COSMOSM Basic FEA System

2-129

Part 2 Verification Problems

TYPE:

Static analysis, PLANE2D element,
coupled degrees of freedom.

PROBLEM:

Determine displacements for the plate
shown in the figure below such that
translations in the Y-direction are
coupled for nodes 5, 10, and 15.

GIVEN:

EX = 3.0E10, 3.0E09, and 3.0E8 psi
ν

= 0.25

COMPARISON OF RESULTS:

Displacements for the coupled D.O.F.

ANALYTICAL SOLUTION:

U10 = U5 = FL/AE

S86: Plate with Coupled Degrees of Freedom

See

List

Young’s

Modulus

U5 (Y-

Translation

at Node 5)

U10 (Y-

Translation
at Node 10)

U15 (Y-

Translation
at Node 15)

U10/U5

Theory

COSMOSM

3.0E10

5.33333E-7

1.000

Theory

COSMOSM

3.0E09

5.33333E-6

1.000

Theory

COSMOSM

3.0E08

5.33333E-5

1.000

Figure S86-1

20 in

10 in

Chapter 2 Linear Static Analysis

2-130

COSMOSM Basic FEA System

TYPE:

Linear static analysis, ELBOW element with pipe cross-section subjected to gravity
loading.

PROBLEM:

Case A:

Reduced gravity loading (fixed-end moments ignored)

Case B:

Consistent gravity loading (fixed-end moments considered)

GIVEN:

= -32.2 in/sec

EX = 3.0E7 psi
ρ

= 7.82

Elbow wall thickness

= 0.1 in

Elbow outer diameter

= 1.0 in

Elbow radius of curvature

= 10.0 in

Figure S87-1

COMPARISON OF RESULTS:

S87: Gravity Loading of ELBOW Element

See

List

Y-Translation at Node 51

Case A

-7.42E-3

Case B

-6.41E-3

Node 51

Node 1

COSMOSM Basic FEA System

2-131

Part 2 Verification Problems

TYPE:

Static analysis, J-integral, stress intensity factor, plane stress conditions.

S88A:

Using 6-node triangular plane element (TRIANG)

S88B:

Using 8-node rectangular plane element (PLANE2D)

REFERENCE:

Brown, W. F., Jr., and Srawley, J. E., “Plane Strain Crack Toughness Testing of High
Strength Metallic Materials,” ASTM Special Technical Publication 410,
Philadelphia, PA, 1966.

PROBLEM:

Determine the stress intensity factor for a single-edge-cracked bend specimen using
the J-integral.

GIVEN:

= 30 x 10

psi

υ = 0.3
Thickness = 1 in
a

= 2 in

= 4 in

= 32 in

P = 1 lb

MODELING HINTS:

Three circular J-integral paths centered at the crack tip are considered. Due to
symmetry, only one half of the model is modeled.

S88A, S88B: Single-Edge Cracked Bend

Specimen, Evaluation of Stress Intensity

Factor Using the J-integral

See

List

Chapter 2 Linear Static Analysis

2-132

COSMOSM Basic FEA System

COMPARISON OF RESULTS

Figure S88-1

Figure S88-2

(TRIANG)

(PLANE2D)

Theory

10.663

Path 1

9.9544

9.1692

Path 2

10.145

10.974

Path 3

10.240

10.648

Full Model

COSMOSM Basic FEA System

2-133

Part 2 Verification Problems

TYPE:

Static analysis, J-integral, stress intensity factors (combined mode crack), plane
strain conditions.

S89A:

Using 6-node triangular plane element (TRIANG)

S89B:

Using 3-node triangular plane element (TRIANG)

S89C:

Using 8-node rectangular plane element (PLANE2D)

S89D:

Using 4-node rectangular plane element (PLANE2D)

REFERENCE:

Bowie, O. L., “Solutions of Plane Crack Problems by Mapping Techniques,” in
Mechanics of Fracture I, Methods of Analysis and Solutions of Crack Problems (Ed
G.C. Shi), pp. 1-55, Noordhoff, Leyden, Netherlands, 1973.

PROBLEM:

Determine the stress intensity factor for both modes of fracture (opening and
shearing) for a rectangular plate with an inclined edge crack subjected to uniform
uniaxial tensile pressure at the two ends.

GIVEN:

σ = 1 psi
h

= 2.5 in

W = 2.5 in
a

= 1 in

= 30 x 10

psi

= 0.3

Thickness =1 in
φ =

MODELING HINTS:

The full part has to be modeled since the model is not symmetric with respect to the
crack. There is no restriction in the type of the mesh to be used and the mesh could

S89A, S89B: Slant-Edge Cracked Plate, Evaluation

of Stress Intensity Factors Using the J-integral

See

List

Chapter 2 Linear Static Analysis

2-134

COSMOSM Basic FEA System

be either symmetric or non-symmetric with respect to the crack. However, the nodes
in the two sides of crack should not be merged in order to model the rupture area
properly.

COMPARISON OF RESULTS:

Reference

1.85

0.880

6-node element

(S89A.GEO)

Path 1
Path 2

1.82
1.82

0.876
0.877

3-node element

(S89B.GEO)

Path 1
Path 2

1.76
1.77

0.835
0.873

8-node element

(S89C.GEO)

Path 1
Path 2

1.80
1.79

0.872
0.874

4-node element

(S89D.GEO)

Path 1
Path 2

1.73
1.71

0.879
0.845

Figure S89-1

Figure S89-2

COSMOSM Basic FEA System

2-135

Part 2 Verification Problems

TYPE:

Static analysis, J-integral, stress intensity factor, axisymmetric geometry.

S90A:

Using 8-node rectangular plane element (PLANE2D)

S90B:

Using 6-node triangular plane element (TRIANG)

REFERENCE:

Tada, H. and Irwin, R., “the stress analysis of cracks Handbook,” Paris Productions,
Inc., pp. 27.1, St. Louis, MI, 1985.

PROBLEM:

Determine the stress intensity factor for a circular crack inside a round bar subjected
to uniform axial tensile pressure at the two ends.

GIVEN:

σ = 1 psi
H = 25 in
R = 5 in
a = 2.5 in
E = 30 x 10

psi

γ = 0.28

MODELING HINTS:

Since the model is symmetric with respect to the crack, therefore only one-half of
the model (lower half here) is needed for the analysis.

S90A, S90B: Penny-Shaped Crack in Round

Bar, Evaluation of Stress Intensity Factor

Using the J-integral

See

List

Chapter 2 Linear Static Analysis

2-136

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

Reference

1.94

8-node element

(S90A.GEO)

Path 1
Path 2

1.90
1.91

6-node element

(S90B.GEO)

Path 1
Path 2

1.89
1.90

Figure S90-1

Figure S90-2

COSMOSM Basic FEA System

2-137

Part 2 Verification Problems

TYPE:

Static analysis, thermal loading, J-integral, stress intensity factor, plane strain
conditions.

REFERENCE:

Wilson, W. K. and Yu, I. W., “The Use of the J-integral in Thermal Stress Crack
Problems,” International Journal of Fracture, Vol. 15, No. 4, August 1979.

PROBLEM:

Determine the stress intensity factor for an edge crack strip subjected to thermal
loading. The strip is subjected to a linearly varying temperature through its thickness
with zero temperature at midthickness and temperature T

at the right edge (x=w/2).

The ends are constrained.

GIVEN:

= 20 in

w = 10 in
a

= 5 in

= 30 x 10

psi

= 0.28

α = 7.4 x 10

-6

in/in-

°F

= 10

°F

MODELING HINTS:

Due to symmetry, only one-half of the geometry is modeled (lower half in this
problem).

COMPARISON OF RESULTS:

where:

β = K

/12220.27

S91: Crack Under Thermal Stresses, Evaluation

of Stress Intensity Using the J-integral

See

List

Chapter 2 Linear Static Analysis

2-138

COSMOSM Basic FEA System

Reference

0.5036

Path 1
Path 2

6141.4
6176.3

0.5034
0.5054

Average value of the five paths in the reference

Figure S91-2

Thickness

Temperature

-To

Figure S91-1

COSMOSM Basic FEA System

2-139

Part 2 Verification Problems

TYPE:

Static analysis, direct material input, SHELL3L element.

REFERENCE:

Timoshenko, S. P. and Woinowsky-Krieger, “Theory of Plates and Shells,”
McGraw-Hill Book Co., 2nd edition, pp. 143-120, 1962.

PROBLEM:

Calculate the deflection and stresses at the center of a simply supported plate
subjected to a concentrated load F.

GIVEN:

MODELING HINTS:

Instead of specifying the elastic properties by E and

ν, the elastic matrix [D] shown

below (in the default element coordinate system) is provided by direct input of its
non-zero terms.

where, [D] relates the element strains to the element stresses according to Hook's
law:

Note that the [D] matrix is reduced to a 5x5 matrix from the general form of 6x6
matrix, by considering the fact that

= 0 for shell element, thus eliminating the

third row and column of the general [D] matrix.

S92A, S92B: Simply Supported Rectangular

Plate, Using Direct Material Matrix Input

See

List

= 30 x 10

psi

= G

= 11.538 x 10

psi

= 0.3

= 1 in

= b = 40 in

= 400 lbs

Chapter 2 Linear Static Analysis

2-140

COSMOSM Basic FEA System

Considering an isotropic property, the terms of [D] matrix are:

The terms K

and K

are

shear correction factors
which are chosen to match
the plate theory with certain
classical solutions and are
functions of thickness and
material properties. When
you input regular material
properties (E,

ν), the shear

factors are evaluated
internally in the program as
K

= K

= 0.1005 (as in

S92B). For the sake of
consistency, the same values
are used for the evaluation of
MC55 and MC66 in S92A.

Due to symmetry in geometry and load, only a quarter of the plate is modeled.

COMPARISON OF RESULTS:

Maximum displacement (in Z-direction) at the tip of the plate (Node 25) using E and

ν (S92B) is compared with the result obtained from direct input of the elastic
coefficients in matrix [D] (S92A).

Theory

Using Direct Matrix

Input (S92A)

Using E and

(S92B)

Maximum UZ

(in)

-0.0270

-0.02746

Figure S92-1

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

2-141

Part 2 Verification Problems

TYPE:

Static analysis, inertia relief, PLANE2D element, (axisymmetric option).

PROBLEM:

A cylinder is accelerating under unbalanced external loads. Find the induced
counter-balance acceleration and the amount by which the cylinder will be
shortened.

GIVEN:

EX = 3.E7 psi
γ

= 0.28 lb sec

/in

= 7.3E-4

Figure S93-1

S93: Accelerating Rocket

See

List

Actual Model

H = 100 in

P = 100 psi

Finite Element Model

R = 10 in

2-142

COSMOSM Basic FEA System

MODELING HINTS:

To avoid instability in FEA solution, one node should be constrained in Y-direction.
A node on the top end of the cylinder is selected for that purpose rather than on the
bottom end. Constraining any node on the surface where the pressure is applied
eliminates the components of the load of that node and hence causes inaccuracy in
the solution.

ANALYTICAL SOLUTION:

a) The induced counter-balance acceleration:

F + Ma = 0

π R

= -

ρ π R

b) Length shortening

COMPARISON OF RESULTS:

* See output file.

Acceleration

(a)

Displacement u

at Node 5

Theory

-1370

0.0001667

COSMOSM

-1370

0.0001669

Figure S93-2

COSMOSM Basic FEA System

2-143

Part 2 Verification Problems

TYPE:

Static analysis using the p-method. S94A: plane stress triangular elements
(TRIANG). S94B: plane stress quadrilateral elements ( PLANE2D). S94C:
Tetrahedral elements (TETRA10).

PROBLEM:

Calculate the maximum stress of a plate with a circular hole under a uniformly
distributed tension load. Use strain energy to adapt the p-order.

GIVEN:

Geometric Properties:

L = side of the plate = 10.00 in

d = diameter of the hole = 1.00 in

t = thickness of the plate = 0.25 in

Material Properties:

E = 3.0E7 psi

ν = 0.3

Loading:

P = 100 psi

A coarse mesh is intentionally used to demonstrate the power of the p-method

S94A, S94B, S94C: P-Method Solution of a Square

Plate with a Small Hole

10”

1” diameter

P
r
e
s

s
u
r
e

stress

concentration

Figure S94-1: The Plate with a Hole Model

Chapter 2 Linear Static Analysis

2-144

COSMOSM Basic FEA System

Figure S94-2 Meshed Quarter of the Plate.

S94C: TETRA10 Elements

S94A: TRIANG Elements

S94B: PLANE2D elements

Convergence Plots Using Different Element Types

COSMOSM Basic FEA System

2-145

Part 2 Verification Problems

COMPARISON OF RESULTS:

Reference:

Walter D. Pilkey, “Formulas For Stress, Strain, and Structural Matrices,” Wiley-
Interscience Publication, John Wiley & Sons, Inc., 1994, pp. 271.

Theory

COSMOSM

Relative Error

Max. Stress in X-

Direction

(TRIANG)

300 psi

308 psi (p-order = 4)

2.7%

Max. Stress in X-

Direction

(PLANE2D)

300 psi

308 psi (p-order = 3)

2.7%

Max. Stress in X-

Direction

(TETRA10)

300 psi

323 psi (p-order = 5)

7.7%

Chapter 2 Linear Static Analysis

2-146

COSMOSM Basic FEA System

TYPE:

Static analysis, axisymmetric triangular (6-node TRIANG) and quadrilateral (8-
node PLANE2D) p-elements with the polynomial order of shape function equal to 8.

PROBLEM:

Calculate the maximum stress of a circular shaft with a U-shape circumferential
groove under a uniformly distributed tension load. P-order is adapted by checking
strain energy of the system.

GIVEN:

Geometric Properties:

L = 0.9 in

D = 2 in

d = 0.2 in

Material Properties:

E = 3.0E7 psi

ν = 0.3

Loading:

P = 100 psi

S95A, S95B, S95C: P-Method Solution of a U-
Shaped Circumferential Groove in a Circular Shaft

Figure S95-1: The Circular Shaft Model

COSMOSM Basic FEA System

2-147

Part 2 Verification Problems

Figure S95-2: Finite Element Model with Different Element Types

S95A: TRIANG elements

S95B: PLANE2D elements

S95C: TETRA10 elements

Convergence Plots Using Different Element Types

Chapter 2 Linear Static Analysis

2-148

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

REFERENCE:

Walter D. Pilkey, “Formulas For Stress, Strain, and Structural Matrices,” Wiley-
Interscience Publication, JohnWiley & Sons, Inc., 1994, pp. 267.

Theory

COSMOSM

Relative Error

Max. Stress in Y-

Direction

(TRIANG)

305 psi

337 psi (p-order = 4 )

10.5%

Max. Stress in Y-

Direction

(PLANE2D)

305 psi

333 psi (p-order = 8)

9.2%

Max. Stress in Y-

Direction

(TETRA10)

305 psi

339 psi (p-order = 5)

10.5%

COSMOSM Basic FEA System

3-1

Modal (Frequency)
Analysis

Introduction

This chapter contains verification problems to demonstrate the accuracy of the
Modal Analysis module DSTAR.

List of Natural Frequency Verification Problems

F1:Natural Frequencies of a Two-Mass Spring System

3-3

F2: Frequencies of a Cantilever Beam

3-4

F3: Frequency of a Simply Supported Beam

3-5

F4: Natural Frequencies of a Cantilever Beam

3-6

F5: Frequency of a Cantilever Beam with Lumped Mass

3-7

F6: Dynamic Analysis of a 3D Structure

3-8

F7A, F7B: Dynamic Analysis of a Simply Supported Plate

3-9

F8: Clamped Circular Plate

3-10

F9: Frequencies of a Cylindrical Shell

3-11

F10: Symmetric Modes and Natural Frequencies of a Ring

3-12

F11A, F11B: Eigenvalues of a Triangular Wing

3-13

F12: Vibration of an Unsupported Beam

3-14

Chapter 3 Modal (Frequency) Analysis

3-2

COSMOSM Basic FEA System

List of Natural Frequency Verification Problems (Concluded)

F13: Frequencies of a Solid Cantilever Beam

3-15

F14: Natural Frequency of Fluid

3-16

F16A, F16B: Vibration of a Clamped Wedge

3-17

F17: Lateral Vibration of an Axially Loaded Bar

3-19

F18: Simply Supported Rectangular Plate

3-20

F19: Lowest Frequencies of Clamped Cylindrical Shell for

Harmonic No. = 6

3-21

F20A, F20B, F20C, F20D, F20E, F20F, F20G, F20H: Dynamic

Analysis of Cantilever Beam

3-22

F21: Frequency Analysis of a Right Circular Canal of Fluid

with Variable Depth

3-23

F22: Frequency Analysis of a Rectangular Tank of Fluid with

Variable Depth

3-25

F23: Natural Frequency of Fluid in a Manometer

3-27

F24: Modal Analysis of a Piezoelectric Cantilever

3-29

F25: Frequency Analysis of a Stretched Circular Membrane

3-30

F26: Frequency Analysis of a Spherical Shell

3-31

F27A, F27B: Natural Frequencies of a Simply-Supported

Square Plate

3-32

F28: Cylindrical Roof Shell

3-33

F29: Frequency Analysis of a Spinning Blade

3-34

COSMOSM Basic FEA System

3-3

Part 2 Verification Problems

TYPE:

Mode shape and frequency, truss and mass element (TRUSS3D, MASS).

REFERENCES:

Thomson, W. T., “Vibration Theory and Application,” Prentice-Hall, Inc.,
Englewood Cliffs, New Jersey, 2nd printing, 1965, p. 163.

PROBLEM:

Determine the normal modes and natural frequencies of the system shown below for
the values of the masses and the springs given.

MODELING HINTS:

Truss elements with zero density are used as springs. Two dynamic degrees of
freedom are selected at nodes 2 and 3 and masses are input as concentrated masses
at nodes 2 and 3.

Figure F1-1

F1: Natural Frequencies of a

Two-Mass Spring System

See

List

GIVEN:

= 2m

= 1 lb-sec

/in

= k

= 200 lb/in

= 4k

= 800 lb/in

COMPARISON OF RESULTS:

, Hz

Theory

2.581

8.326

COSMOSM

2.581

8.326

Problem Sketch

1st

D.O.F.

2nd

D.O.F.

Finite Element Model

Chapter 3 Modal (Frequency) Analysis

3-4

COSMOSM Basic FEA System

TYPE:

Mode shape and frequency, plane element (PLANE2D).

REFERENCE:

Flugge, W., “Handbook of Engineering Mechanics,” McGraw-Hill Book Co.,

Inc.,

New York, 1962, pp. 61-6, 61-9.

PROBLEM:

Determine the fundamental frequency, f, of the cantilever beam of uniform cross
section A.

Figure F2-1

F2: Frequencies of a Cantilever Beam

See

List

GIVEN:

= 30 x 10

psi

= 50 in

= 0.9 in

= 0.81 in

= 0

= 0.734E-3 lb sec

/in

COMPARISON OF RESULTS

, Hz

Theory

11.79

74.47

208.54

COSMOSM

11.72

73.35

206.68

Finite Element Model

Problem Sketch

Front View

Cross Section

COSMOSM Basic FEA System

3-5

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, beam element (BEAM3D).

REFERENCE:

Thomson, W. T., “Vibration Theory and Applications,” Prentice-Hall, Inc.,
Englewood Cliffs, New Jersey, 2nd printing, 1965, p. 18.

PROBLEM:

Determine the fundamental frequency, f, of the simply supported beam of uniform
cross section A.

GIVEN:

= 30 x 10

psi

= 80 in

= 0.7272E-3 lb-sec

/in

= 4 in

= 1.3333 in

= 2 in

ANALYTICAL
SOLUTION:

π)

(EI/mL

)

(1/2)

= Number of frequencies

COMPARISON OF RESULTS:

F3: Frequency of a Simply Supported Beam

See

List

, Hz

Theory

28.78

115.12

259.0

COSMOSM

28.78

114.31

242.7

Figure F3-1

Finite Element Model

Problem Sketch

Chapter 3 Modal (Frequency) Analysis

3-6

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, beam element (BEAM3D).

REFERENCE:

Thomson, W. T., “Vibration Theory and Applications,” Prentice-Hall, Inc.,
Englewood Cliffs, New Jersey, 2nd printing, 1965, p. 278, Ex. 8.5-1, and p. 357.

PROBLEM:

Determine the first three
natural frequencies, f, of a
uniform beam clamped at
one end and free at the
other end.

GIVEN:

= 30 x 10

psi

I =

1.3333

= 4 in

= 2 in

= 80 in

= 0.72723E-3 lb sec

/in

COMPARISON OF RESULTS:

F4: Natural Frequencies of a Cantilever Beam

See

List

, Hz

Theory

10.25

64.25

179.9

COSMOSM

10.24

63.95

178.5

Problem Sketch

1 2 3 4

1 2

Finite Element Model

Figure F4-1

COSMOSM Basic FEA System

3-7

Part 2 Verification Problems

TYPE:

Mode shape and frequency, beam and mass elements (BEAM3D, MASS).

REFERENCE:

William, W. Seto, “Theory and Problems of Mechanical Vibrations,” Schaum’s
Outline Series, McGraw-Hill Book Co., Inc., New York, 1964, p. 7.

PROBLEM:

A steel cantilever beam of
length 10 in has a square cross-
section of 1/4 x 1/4 inch. A
weight of 10 lbs is attached to
the free end of the beam as
shown in the figure. Determine
the natural frequency of the
system if the mass is displaced
slightly and released.

GIVEN:

= 30 x 10

psi

W = 10 lb

= 10 in

COMPARISON OF RESULTS:

F5: Frequency of a Cantilever Beam

with Lumped Mass

See

List

F, Hz

Theory

5.355

COSMOSM

5.359

Problem Sketch

Finite Element Model

Figure F5-1

Chapter 3 Modal (Frequency) Analysis

3-8

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, pipe and mass elements (PIPE, MASS).

REFERENCE:

“ASME Pressure Vessel and
Piping 1972 Computer
Programs Verification,” Ed.
by I. S. Tuba and W. B.
Wright, ASME Publication
I-24, Problem 1.

PROBLEM:

Find the natural frequencies
and mode shapes of the 3D
structure given below.

GIVEN:

Each member is a pipe.

Outer diameter = 2.375 in

Thickness = 0.154 in

= 27.9 x 10

psi

= 0.3

The masses are represented solely by lumped masses as shown in the figure.

= M

= 0.00894223 lb sec

/in

= M

= 0.0253816 lb sec

/in

COMPARISON OF RESULTS:

F6: Dynamic Analysis of a 3D Structure

See

List

, Hz

Theory

111.5

115.9

137.6

218.0

404.2

COSMOSM

111.2

115.8

137.1

215.7

404.2

8.625

17.25

18.625

10.0

18.625

27.25

Problem Sketch and Finite Element Model

Figure F6-1

COSMOSM Basic FEA System

3-9

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, shell elements (SHELL4 and SHELL6).

REFERENCE:

Leissa, A.W. “Vibration of Plates,” NASA, sp-160, p. 44.

PROBLEM:

Obtain the first natural
frequency for a simply
supported plate.

GIVEN:

= 30,000 kips

= 0.3

= 1 in

= b = 40 in

= 0.003 kips sec

/in

NOTE:

Due to double symmetry in geometry and the required mode shape, a quarter of the
plate is taken for modeling.

COMPARISON OF RESULTS

The first natural frequency of the plate is 5.94 Hz.

F7A, F7B: Dynamic Analysis of a Simply

Supported Plate

See

List

F7A: SHELL4

F7B: SHELL6

(Curved)

F7B: SHELL6

(Assembled)

COSMOSM

5.93 Hz

5.94 Hz

5.93

Problem Sketch and Finite Element Model

Figure F7-1

Chapter 3 Modal (Frequency) Analysis

3-10

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, thick shell element (SHELL3T).

REFERENCE:

Leissa A.W., “Vibration of Plates,” NASA sp-160, p. 8.

PROBLEM:

Obtain the first three
natural frequencies.

GIVEN:

E = 30 x 10

psi

ν = 0.3
ρ = 0.00073 (lb/in

) sec

R = 40 in

t = 1 in

NOTE:

Since a quarter of the plate is used for modeling, the second natural frequency is not
symmetric (s = 0, n = 1) and will not be calculated. This is an example to show that
symmetry should be used carefully.

COMPARISON OF RESULTS:

F8: Clamped Circular Plate

See

List

Frequency No.

Theory (Hz)

COSMOSM (Hz)

62.30

62.40

212.60

212.53

242.75

240.30

s* refers to the number of nodal circles

n* refers to the number of nodal diameters

Finite Element Model

Problem Sketch

Figure F8-1

COSMOSM Basic FEA System

3-11

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, shell element (SHELL4).

REFERENCE:

Kraus, “Thin Elastic Shells,” John Wiley & Sons, Inc., p. 307.

PROBLEM:

Determining the first three
natural frequencies.

GIVEN:

= 30 x 10

psi

= 0.3

= 0.00073 (lb-sec

)/in

= 12 in

= 3 in

= 0.01 in

NOTE:

Due to symmetry in geometry and the mode shapes of the first three natural
frequencies, 1/8 of the cylinder is considered for modeling.

COMPARISON OF RESULTS:

F9: Frequencies of a Cylindrical Shell

See

List

, Hz

Theory

552

736

783

COSMOSM

553.69

718.50

795.60

Problem Sketch

and Finite Element Model

Figure F9-1

Chapter 3 Modal (Frequency) Analysis

3-12

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, shell element (SHELL4).

REFERENCE:

Flugge, W. “Handbook of
Engineering Mechanics,” First
Edition, McGraw-Hill, New York,
p. 61-19.

PROBLEM:

Determine the first two natural
frequencies of a uniform ring in
symmetric case.

GIVEN:

= 30E6 psi

= 0

= 4 in

= 1 in

= 0.25E-2 (lb sec

)/in

COMPARISON OF RESULTS:

F10: Symmetric Modes and Natural

Frequencies of a Ring

See

List

, Hz

Theory

135.05

735.14

COSMOSM

134.92

723.94

Problem Sketch

Figure F10-1

COSMOSM Basic FEA System

3-13

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, triangular shell elements (SHELL3 and SHELL6).

REFERENCE:

“ASME Pressure Vessel and Piping 1972 Computer Programs Verification,” ed. by
I. S. Tuba and W. B. Wright, ASME Publication I-24, Problem 2.

PROBLEM:

Calculate the natural
frequencies of a triangular
wing as shown in the figure.

GIVEN:

= 6.5 x 10

psi

= 0.3541

= 0.166E-3 lb sec

/in

= 6 in

Thickness = 0.034 in

COMPARISON OF RESULTS:

Natural Frequencies (Hz):

F11A, F11B: Eigenvalues of a Triangular Wing

See

List

Frequency

No.

Reference

COSMOSM

SHELL3

SHELL6

(Curved)

SHELL6

(Assembled)

55.9

55.8

56.137

55.898

210.9

206.5

212.708

210.225

293.5

285.5

299.303

291.407

Finite Element Model

Problem Geometry

Figure F11-1

Chapter 3 Modal (Frequency) Analysis

3-14

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, rigid body modes, beam element (BEAM3D).

REFERENCE:

Timoshenko, S. P., Young, O. H., and Weaver, W., “Vibration Problems in
Engineering,” 4th ed., John Wiley and Sons, New York, 1974, pp. 424-425.

PROBLEM:

Determine the elastic and
rigid body modes of vibration
of the unsupported beam
shown below.

GIVEN:

= 100 in

= 1 x 10

psi

= 0.1 in

= 0.2588E-3 lb sec

/in

ANALYTICAL SOLUTION:

The theoretical solution is given by the roots of the equation Cos KL Cosh KL = 1
and the frequencies are given by:

COMPARISON OF RESULTS:

NOTE:

First two modes are rigid body modes.

F12: Vibration of an Unsupported Beam

See

List

= K

(EI/

ρA)

(1/2)

/(2

π)

= Number of natural frequencies

= (i + 0.5)

π/L

= area of cross-section

= Mass Density

Mode 1

Mode 2

Mode 3

Mode 4

Mode 5

Mode 6

Theory F, Hz

11.07

30.51

59.81

98.86

Theory (ki)

(0)

(4.73)

(7.853)

(10.996) (14.137)

COSMOSM F, Hz

10.92

29.82

57.94

94.94

Figure F12-1

15 16

Finite Element Model

Problem Sketch

COSMOSM Basic FEA System

3-15

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, hexahedral solid element (SOLID).

REFERENCE:

Thomson, W. T., “Vibration Theory and Applications,” Prentice-Hall, Inc.,
Englewood Cliffs, N. J., 2nd printing, 1965, p.275, Ex. 8.5-1, and p. 357.

PROBLEM:

Determine the first
three natural
frequencies of a
uniform beam
clamped at one
end and free at
the other end.

GIVEN:

= 30 x 10

psi

= 2 in

= 80 in

= 0.00072723
lb-sec

/in

COMPARISON OF RESULTS:

F13: Frequencies of a Solid Cantilever Beam

See

List

, Hz

Theory

10.25

64.25

179.91

COSMOSM

10.24

63.95

178.38

Problem Sketch

Finite Element

Model

Figure F13-1

Chapter 3 Modal (Frequency) Analysis

3-16

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, truss elements (TRUSS2D).

REFERENCE:

William,

Seto,

“Theory and Problems of Mechanical

Vibrations,”

Schaum’s

Outline Series, McGraw-Hill Book Co., Inc., New York, 1964, p. 7.

PROBLEM:

A manometer used in a fluid mechanics laboratory has a uniform bore of cross-
section area A. If a column of liquid of length L and weight density

is set into

motion, as shown in the figure, find the frequency of the resulting motion.

NOTE:

The mass of fluid is lumped at nodes 2 to 28. The boundary elements are applied at
nodes 6 to 24.

Figure F14-1

F14: Natural Frequency of Fluid

See

List

GIVEN:

COMPARISON OF RESULTS:

= 1 in

ρ = 9.614E-5 lb sec

/in

= 51.4159 in

= 1E5 psi

F, Hz

Theory

0.617

COSMOSM

0.617

Problem Sketch

Finite Element Model

1.0"

10"

COSMOSM Basic FEA System

3-17

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, thick shell elements (SHELL3T, SHELL4T).

REFERENCE:

Timoshenko, S., and Young, D. H., “Vibration Problems in Engineering,” 3rd
Edition, D. Van Nostrand Co., Inc., New York, 1955, p. 392.

PROBLEM:

Determine the fundamental frequency of lateral vibration of a wedge shaped plate.
The plate is of uniform thickness t, base 3b, and length L.

GIVEN:

= 30 x10

psi

= 7.28E-4 lb sec

/in

= 1 in

= 2 in

= 16 in

MODELING HINTS:

Only in-plane (in x-y plane) frequencies along y-direction are considered. In order
to find better results, out-of-plane displacements (z-direction) are restricted.

The effect of different elements and meshes is also considered.

ANALYTICAL SOLUTION:

The first in-plane natural frequency calculated by:

Using approximate RITZ method, for first and second natural frequencies:

F16A, F16B: Vibration of a Clamped Wedge

See

List

Chapter 3 Modal (Frequency) Analysis

3-18

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

Figure F16A-1

Natural Frequency (Hz)

First

Second

Reference
Exact
Ritz

774.547
775.130

---

2521.265

COSMOSM
SHELL3T (F16A)
SHELL4T (F16B)

813.45
789.12

2280.78
2309.54

Side View

Vibration
Direction

Fine Mesh SHELL4T

Elements

Coarse Mesh SHELL3T

Elements

Problem Sketch

Plan

COSMOSM Basic FEA System

3-19

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, in-plane effects, beam elements (BEAM3D).

REFERENCE:

Timoshenko, S., and Young, D. H., “Vibration Problems in Engineering,” 3rd
Edition, D. Van Nostrand Co., Inc., New York, 1955, p. 374.

PROBLEM:

Determine the fundamental frequency of lateral vibration of a wedge shaped plate.
The plate is of uniform thickness t, base 3b, and length L.

GIVEN:

COMPARISON OF RESULTS:

Figure F17-1

F17: Lateral Vibration of an Axially Loaded Bar

See

List

= 30E6 psi

= 7.2792E-4 lb sec

/in

= 386 in/sec

= h = 2 in

= 80 in

= 40,000 lb

, Hz

Theory

17.055

105.32

249.39

COSMOSM

17.055

105.32

249.34

Problem Sketch

FInite Element Model

Chapter 3 Modal (Frequency) Analysis

3-20

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, in-plane effects, shell element (SHELL4).

REFERENCE:

Leissa, A.W., “Vibration of Plates,” NASA, p-160, p. 277.

PROBLEM:

Obtain the fundamental frequency of a simply supported plate with the effect of in-
plane forces. Nx = 33.89 lb/in applied at x = 0 and x = a.

NOTE:

Due to double symmetry in geometry, loads and the mode shape, a quarter plate is
taken for modeling.

Figure F18-1

F18: Simply Supported Rectangular Plate

See

List

GIVEN:

COMPARISON OF RESULTS:

= 30,000 psi

ν = 0.3
h

= 1 in

= b = 40 in

= 0.0003 (lb sec

)/in

= 33.89 psi

F, Hz

Theory

4.20

COSMOSM

4.19

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

3-21

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, axisymmetric shell elements (SHELLAX).

REFERENCE:

Leissa, A. W., “Vibration of Shells,” NASA sp-288, p. 92-93 (1973).

PROBLEM:

To find the lowest natural
frequency of vibration for the
cylinder fixed at both ends.

GIVEN:

= 3 in

= 12 in

= 0.01 in

= 30 x 10

psi

= 0.35

= 0.000730 lb sec

/in

Range of circumferential
harmonics (n) = 4 to 7

MODELING HINTS:

All the 21 nodes are spaced equally along the meridian of cylinder. The number of
circumferential harmonics (lobes) for each frequency analysis is to be specified and
lowest frequency is sought.

COMPARISON OF RESULTS:

F19: Lowest Frequencies of Clamped

Cylindrical Shell for Harmonic No. = 6

See

List

Harmonic No.

(n)

First Frequency (Hz)

Theory

Experiment

COSMOSM

(4) *

926

700

777.45

(5) *

646

522

592.6

(6) *

563

525

549.4

(7) *

606

592

609.7

* You need to re-execute the analysis by specifying these harmonic

numbers under the A_FREQUENCY command. The lowest natural
frequency is 549.6 Hz corresponding to harmonic number = 6.

Finite Element Model

Problem Sketch

Figure F19-1

Chapter 3 Modal (Frequency) Analysis

3-22

COSMOSM Basic FEA System

TYPE:

Mode shapes and frequencies, multifield elements, 4- and 8-node PLANE2D,
SHELL4T, 6-node TRIANG, TETRA10, 8- and 20-node SOLID, TETRA4R, and
SHELL6.

PROBLEM:

Compare the first two natural frequencies of a cantilever beam modeled by each of
the above element types.

GIVEN:

= 10

psi

= 245 x 10

–3

lb-sec

/in

= 0.1 in

= 0.2 in

= 6 in

= 0.3

COMPARISON OF RESULTS:

The theoretical solutions for the first and second mode are: 181.17 and 1136.29 Hz.

F20A, F20B, F20C, F20D, F20E, F20F, F20G, F20H:

Dynamic Analysis of Cantilever Beam

See

List

Input

File

Element

1st Mode

Error (%)

2nd Mode

Error (%)

F20A

PLANE2D 4-node

180.71

0.2

1127.96

0.7

F20B

PLANE2D 8-node

181.15

0.0

1153.52

1.53

F20C

TRIANG 6-node

183.35

1.2

1182.90

4.1

F20D

TETRA10

183.10

1.0

1184.85

4.3

F20E

SOLID 8-node

181.64

0.2

1134.67

0.2

F20F

SOLID 20-node

179.72

0.8

1111.16

2.2

F20G

TETRA4R

190.24

5.1

1182.72

4.1

F20H

SHELL6 (Curved)

183.371

1.2

1182.87

4.1

SHELL6
(Assembled)

183.357

1.2

1182.54

4.1

Problem Sketch

Figure F20-1

COSMOSM Basic FEA System

3-23

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, fluid sloshing, plane strain elements (PLANE2D).

REFERENCE:

Budiansky, B., “Sloshing of Liquids in Circular Canals and Spherical Tank,” J.
Aerospace Sci, 27, p. 161-173, (1960).

PROBLEM:

A right circular canal with radius R is half-filled by an incompressible liquid (see
Figure F21-1). Determine the first two natural frequencies with mode shapes
antisymmetric about the Y-axis.

GIVEN

= 56.4 in

H/R = 0
ρ = 0.9345E-4 lb sec

/in

EX = 3E5 lb/in

Where:

EX = bulk modulus

NOTES:

A small shear modulus SXY = EX is used to prevent numerical instability.

The radial component of displacements (in local cylindrical coordinate system)
is constrained at the curved boundary in order to allow sloshing.

The acceleration due to gravity (ACEL command) in the negative direction.

PLANE2D plane strain elements are used to solve the current problem since the
mode shapes are independent of Z-direction coordinates.

For non rectangular geometries, one can expect to obtain some natural
frequencies with no significant changes in the free surface profile. This situation
is analogous to the rigid modes of a solid structure. Therefore, a negative shift of

is recommended to prevent this type of sloshing modes.

F21: Frequency Analysis of a Right Circular

Canal of Fluid with Variable Depth

See

List

Chapter 3 Modal (Frequency) Analysis

3-24

COSMOSM Basic FEA System

COMPARISON OF RESULTS:

Figure F21-1

Mode

Number

Analytical

Solution

(Hz)

COSMOSM

(Hz)

0.4858

0.4875

Not Available

0.7269

0.9031

0.8976

Free Surface

Constrain only the radial
component of displacement
to allow sloshing.

Finite Element Model

COSMOSM Basic FEA System

3-25

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, fluid sloshing, hexahedral solid (SOLID).

REFERENCE:

Lamb, H., “Hydrodynamics,” 6th edition, Dover Publications, Inc., New York,
1945.

PROBLEM:

A rectangular tank with dimensions A and B in X- and Z-directions is partially filled
by an incompressible liquid (see Figure F22-1). Determine the first two natural
frequencies.

GIVEN:

= 48 in

= 20 in

= 0.9345E-4 lb sec

/in

EX = 3E5 lb/in

Where:

EX = bulk modulus

NOTE:

Please refer to notes (1), (2), (3), (4) and (5) in Problem F21.

COMPARISON OF RESULTS:

The analytical solution for natural frequencies is as follows:

where i and j represent the order number in X- and Z-directions respectively.

F22: Frequency Analysis of a Rectangular

Tank of Fluid with Variable Depth

See

List

Chapter 3 Modal (Frequency) Analysis

3-26

COSMOSM Basic FEA System

The comparison of analytical solutions with those obtained using COSMOSM for
various values of i and j are tabulated below.

Figure F22-1

Frequency

Number

I / J

Analytical

Solution

(Hz)

COSMOSM

(Hz)

1 / 0

—

0 /1

0.7440

0.7422

1 / 2

0.9286

0.9199

Free Surface

Constrain the normal component of
displacement to allow sloshing.

Finite Element Model

COSMOSM Basic FEA System

3-27

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies, fluid sloshing, plane strain elements (PLANE2D).

REFERENCE:

William, W. Seto, “Theory and Problems of Mechanical Vibrations,” Schaum's
Outline Series, McGraw-Hill Book Co., Inc., New York, 1964, p. 7.

PROBLEM:

A manometer used in a fluid mechanics laboratory has a uniform bore of cross-
sectional area A. If a column of liquid of length L and weight density r is set into
motion as shown in the figures, find the frequency of the resulting motion.

NOTE:

A small shear modulus GXY = EX(1.0E-9) is used to prevent numerical instability.
Global and local constraints are applied normal to the boundary to prevent leaking
of the fluid. Acceleration due to gravity (ACEL command) in the negative y-
direction should be included for problems with free surfaces

F23: Natural Frequency of Fluid in a Manometer

See

List

GIVEN:

COMPARISON OF RESULTS:

= 0.5 in

= 0.9345E-4 lb sec

/in

= 26.4934 in (length of
fluid in the manometer)

EX = 3E5 lb/in

Where:

EX = bulk modulus

F, Hz

Analytical Solution *

0.8596

COSMOSM

0.8623

Chapter 3 Modal (Frequency) Analysis

3-28

COSMOSM Basic FEA System

Figure F23-1

Problem Sketch

Finite Element Model

Constrain
displacement
components
normal to the
surface to allow
fluid sloshing

5 in

0.5 in

5 in

Free
Surface

COSMOSM Basic FEA System

3-29

Part 2 Verification Problems

TYPE:

Mode shapes and frequencies using solid piezoelectric element (SOLIDPZ).

REFERENCE:

J. Zelenka, “Piezoelectric Resonators and their Applications”, Elsevier Science
Publishing Co., Inc., New York, 1986.

PROBLEM:

A piezoelectric transducer with a polarization direction along its longitudinal
direction has electrodes at two ends. Both electrodes are grounded to represent a
short-circuit condition. All non-prescribed voltage D.O.F.'s are condensed out after
assemblage of stiffness matrix. In this problem, the longitudinal mode of vibration
is under consideration.

GIVEN:

= 80 mm

= h = 2 mm

Density

= 727 Kg/m

NOTE:

To constrain voltage degrees of
freedom for piezoelectric applica-
tion, use the RX component of
displacement in the applicable constraint commands (DND, DCR, DSF, etc.). There
is no rotational degree of freedom for SOLID elements in COSMOSM.

COMPARISON OF RESULTS:

For the sixth mode of vibration in this problem (longitudinal mode):

F24: Modal Analysis of a Piezoelectric Cantilever

See

List

Sixth Mode of Vibration

(Longitudinal Mode)

Theory

690 Hz

COSMOSM

685 Hz

L = 80 mm
b = 2 mm
h = 2 mm

= 727 kg/cu m

Figure F24-1

Chapter 3 Modal (Frequency) Analysis

3-30

COSMOSM Basic FEA System

TYPE:

Frequency analysis using the nonaxisymmetric mode shape option (SHELLAX).

REFERENCE:

Leissa, A. W., “Vibration of Shells,” NASA-P-SP-288, 1973.

PROBLEM:

Find the first three frequencies of a stretched circular membrane.

MODELING HINTS:

A total of 9 elements are considered as shown. The stretching load of 1500 lb for a
one radian section of the shell is applied with the inplane loading flag turned on for
frequency calculations. All frequencies are found for circumferential harmonic
number 0.

Figure F25-1

F25: Frequency Analysis of a Stretched

Circular Membrane

See

List

GIVEN:

COMPARISON OF RESULTS

= 15 in

= 30E6 psi

= 100 lb/in

= 0.01 in (Thickness)

= 0.0073 lb-sec

/in

Natural

Frequency No.

Theory

(Hz)

COSMOSM

(Hz)

Error

(%)

94.406

93.73

0.72

216.77

212.95

1.76

339.85

329.76

2.97

Finite Element Model

Problem Sketch

COSMOSM Basic FEA System

3-31

Part 2 Verification Problems

TYPE:

Frequency analysis using the nonaxisymmetric mode shape option (SHELLAX).

REFERENCE:

Krause, H., “Thin Elastic Shells,” John Wiley, Inc., New York, 1967.

PROBLEM:

Find the first eight
frequencies of the
spherical shell shown
here for the
circumferential
harmonic number 2.

GIVEN:

= 10 in

= 1E7 psi

= 0.0005208 lb-sec

/in

ν (NUXY) = 0.3
t (Thickness) = 0.1 in

COMPARISON OF RESULTS:

F26: Frequency Analysis of a Spherical Shell

See

List

Natural

Frequency No.

Theory

(Hz)

COSMOSM

(Hz)

Error (%)

1620

1622

0.12

1919

1923

0.20

2035

2044

0.44

2093

2110

0.81

2125

2153

1.32

2145

2188

2.00

2159

2224

3.01

2168

2262

4.34

Finite Element Model

Problem Sketch

Figure F26-1

Chapter 3 Modal (Frequency) Analysis

3-32

COSMOSM Basic FEA System

TYPE:

Frequency analysis, Guyan reduction, SHELL4 elements.

Case A:

Guyan Reduction

Case B:

Consistent Mass

PROBLEM:

Natural frequencies of a simply-supported plate are calculated. Utilizing the
symmetry of the model, only one quarter of the plate is modeled and the first three
symmetric modes of vibration are calculated. The mass is lumped uniformly at
master degrees of freedom.

Theoretical results can be
obtained from the equation:

= r

D / L

∗

+ n

)

Where:

D = Eh

/ 12(1 -

)

U =

ρh

F27A, F27B: Natural Frequencies of a

Simply-Supported Square Plate

See

List

GIVEN:

= 30 in

= 0.1 in

= 8.29 x 10

-4

(lb sec

)/in

= 0.3

= 30.E6 psi

ANALYTICAL
SOLUTION:

COMPARISON OF RESULTS:

Normalized mode shape displacements for the nodes
connected by the rigid bar.

Natural Frequency (Hz)

First

Second

Third

Theory

5.02

25.12

Case A: Guyan Reduction

5.03

25.15

25.20

Case B: Consistent Mass

5.02

25.11

Total Mass =

∗

8.29

∗

-4

∗

0.1

∗

30 =.07461

Lumped Mass at Master Nodes = .07461/64 = 1.16E-3

Problem Sketch

961

931

Simply
Supported
Plate

Figure F27-1

COSMOSM Basic FEA System

3-33

Part 2 Verification Problems

TYPE:

Natural mode shape and frequency, shell and rigid bar elements.

PROBLEM:

Determine the first frequency and mode shape of the shell roof shown below.

GIVEN:

r = 25 ft

E = 4.32E12,

4.32E11, and
4.32E10 psi

ν = 0

MODELING HINTS:

Due to symmetry, a
quarter of the shell roof
is considered in the
modeling. Nodes 8 and
12 are connected by a
rigid bar.

COMPARISON OF RESULTS:

Normalized mode shape displacements for the nodes connected by the rigid bar.

F28: Cylindrical Roof Shell

See

List

Method

Young's

Modulus

Z-Rotation

R8 / R12

Node 8 (R8)

Node 12 (R12)

Theory

COSMOSM

4.32E12

-0.5642901E-2
-0.5720E-2

1.000
1.000

Theory

COSMOSM

4.32E11

-0.5654460E-2
-0.5720E-2

1.000
1.000

Theory

COSMOSM

4.32E10

-0.5693621E-2
-0.5720E-2

1.000
1.000

25 ft

Free Edge

v =
w = 0

t = 0.25 ft

v = w = 0

Problem Sketch and Finite

Element Model

Free Edge

Figure F28-1

Chapter 3 Modal (Frequency) Analysis

3-34

COSMOSM Basic FEA System

TYPE:

Frequency analysis using the Spin Softening and Stress Stiffening Options.

REFERENCE:

W. Carnegie, “Vibrations of Rotating Cantilever Blade,” J. of Mechanical
Engineering Science, Vol. 1, No. 3, 1959.

PROBLEM:

Find the fundamental frequency of vibration of a blade cantilevered from a rigid
spinning rod.

Figure F29-1

MODELING HINTS:

The blade is cantilevered to a rigid rod. Therefore, the blade may be modeled with a
fixed displacement boundary condition at the connection to the rod. The Stress Stiff-
ening effect due to centrifugal load is considered in this model by activating the cen-
trifugal force option in

A_STATIC

command together with the Inplane Loading Flag

A_FREQUENCY

command.

F29A, B, C: Frequency Analysis

of a Spinning Blade

See

List

Actual Model

Finite Element Model

COSMOSM Basic FEA System

3-35

Part 2 Verification Problems

GIVEN:

= 150 mm

E = 217 x 10

= 328 mm

ρ = 7850 Kg/m

= 28 mm

γ = 0.3

= 3 mm

ω = 314.159 rad/sec

COMPARISON OF RESULTS:

Fundamental

Frequency (Hz)

Error (%)

Theory

52.75

A Stress stiffening with spin softening

51.17

3.0

B Stress stiffening with no spin softening

71.54

36.0

C No stress stiffening and no spin softening

23.80

54.9

3-36

COSMOSM Basic FEA System

4-1

Buckling Analysis

Introduction

This chapter contains verification problems to demonstrate the accuracy of the
Buckling Analysis module DSTAR.

List of Buckling Verification Problems

B1: Instability of Columns

4-2

B2: Instability of Columns

4-3

B3: Instability of Columns

4-4

B4: Simply Supported Rectangular Plate

4-5

B5A, B5B: Instability of a Ring

4-6

B6: Buckling Analysis of a Small Frame

4-7

B7A, B7B: Instability of Frames

4-9

B8: Instability of a Cylinder

4-10

B9: Simply Supported Stiffened Plate

4-11

B10: Stability of a Rectangular Frame

4-12

B11: Buckling of a Stepped Column

4-13

B12: Buckling Analysis of a Simply Supported Composite Plate

4-14

B13: Buckling of a Tapered Column

4-15

B14: Buckling of Clamped Cylindrical Shell Under External Pressure

Using the Nonaxisymmetric Buckling Mode Option

4-16

B15A, B15B: Buckling of Simply-Supported Cylindrical Shell Under Axial Load

4-18

Chapter 4 Buckling Analysis

4-2

COSMOSM Basic FEA System

TYPE:

Buckling analysis, beam element (BEAM3D).

REFERENCE:

Brush, D. O., and Almroth, B. O., “Buckling of Bars, Plates, and Shells,” McGraw-
Hill, Inc., New York, 1975, p. 22.

PROBLEM:

Find the buckling load and deflection mode for a simply supported column.

ANALYTICAL SOLUTION:

EI / L

= 9869.6 lb

Figure B1-1

B1: Instability of Columns

See

List

GIVEN:

= 30 x 10

psi

= 1 in

= 50 in

= 1/12 in

COMPARISON OF RESULTS:

Theory

COSMOSM

9869.6 lb

Problem Sketch

1 2 3

Finite Element Model

COSMOSM Basic FEA System

4-3

Part 2 Verification Problems

TYPE:

Buckling analysis, beam element (BEAM3D).

REFERENCE:

Brush, D. O., and Almroth, B. O., “Buckling of Bars, Plates, and Shells,” McGraw-
Hill, Inc., New York, 1975, p. 22.

PROBLEM:

Find the buckling load and deflection mode for a clamped-clamped column.

ANALYTICAL SOLUTION:

= 4

EI / L

= 39478.4 lb

Figure B2-1

B2: Instability of Columns

See

List

GIVEN:

= 30 x 10

psi

= 1 in

= 50 in

= 1/12 in

COMPARISON OF RESULTS:

Theory

COSMOSM

39478.4 lb

39478.8 lb

1 2 3

Problem Sketch

Finite Element Model

Chapter 4 Buckling Analysis

4-4

COSMOSM Basic FEA System

TYPE:

Buckling analysis, beam element (BEAM3D).

REFERENCE:

Brush, D. O., and Almroth, B. O., “Buckling of Bars, Plates, and Shells,” McGraw-
Hill, Inc., New York, 1975, p. 22.

PROBLEM:

Find the buckling load and deflection mode for a clamped-free column.

ANALYTICAL SOLUTION:

EI / (4L

) = 2467.4 lb

Figure B3-1

B3: Instability of Columns

See

List

GIVEN:

= 30 x 10

psi

= 1 in

= 50 in

= 1/12 in

COMPARISON OF RESULTS:

Theory

COSMOSM

2467.4 lb

Finite Element Model

1 2 3

Problem Sketch

COSMOSM Basic FEA System

4-5

Part 2 Verification Problems

TYPE:

Buckling analysis, shell element (SHELL4).

REFERENCE:

Timoshenko, and Woinosky-Krieger, “Theory of Plates and Shells,” McGraw-Hill
Book Co., New York, 2nd Edition, p. 389.

PROBLEM:

Find the buckling load of a simply supported isotropic plate subjected to inplane
uniform load p applied at x = 0 and x = a.

Figure B4-1

B4: Simply Supported Rectangular Plate

See

List

GIVEN:

COMPARISON OF RESULTS:

= 30,000 psi

= 0.3

= 1 in

= b = 40 in

= 1 lb/in

Theory

COSMOSM

67.78 lb

67.85 lb

NOTE:

Due to double symmetry in geometry and loads,
a quarter of the plate is taken for modeling.

Problem Sketch and Finite Element Model

Chapter 4 Buckling Analysis

4-6

COSMOSM Basic FEA System

TYPE:

Buckling analysis, shell element (SHELL3, SHELL6).

REFERENCE:

Brush, D. O., and Almroth, B. O., “Buckling of Bars, Plates, and Shells,” McGraw-
Hill, Inc., New York, 1975, p. 139.

PROBLEM:

Find the buckling load and deflection mode of a ring under pressure loading.

ANALYTICAL SOLUTION:

Using Donnell Approximations.

= 4EI / R

= 26.667 lb/in

Figure B5-1

B5A, B5B: Instability of a Ring

See

List

GIVEN:

= 10 x 10

psi

= 5 in

= 0.1 in

= 1 in

= 0.001/12 in

COMPARISON OF RESULTS:

Theory

COSMOSM

SHELL3

SHELL6

(Curved)

SHELL6

(Assembled)

Pcr

26.667 lb

26.674 lb

26.648 lb

-26.660 lb

Problem Sketch

Finite Element Model

Pcr

COSMOSM Basic FEA System

4-7

Part 2 Verification Problems

TYPE:

Buckling analysis, truss (TRUSS2D) and beam (BEAM3D) elements.

REFERENCE:

Timoshenko, S. P., and Gere, J. M., “Theory of Elastic Stability,” 2nd ed., McGraw-
Hill Book Co., New York, 1961, p. 45.

ANALYTICAL SOLUTION:

The classical results are obtained from:

1cr

= A

E Sin

α Cos

α / (1 + (A

) Sin

α)

2cr

/ L

MODE SHAPES:

Figure B6-1

B6: Buckling Analysis of a Small Frame

See

List

GIVEN:

= 20 in

= 4 in

= 0.1 in

= E

= 30E6 psi

= 2 in

COMPARISON OF RESULTS:

Theory

COSMOSM

1051.392 lb

1051.367 lb

1480.44 lb

1481.20 lb

Mode Shape 1

Mode Shape 2

Chapter 4 Buckling Analysis

4-8

COSMOSM Basic FEA System

Figure B6-2

Truss

= 45

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

4-9

Part 2 Verification Problems

TYPE:

Buckling analysis, shell element (SHELL4 and SHELL6).

REFERENCE:

Brush, D. O., and Almroth, B. O., “Buckling of Bars, Plates, and Shells,” McGraw-
Hill, Inc., New York, 1975, p.29.

PROBLEM:

Find the buckling load and deflection mode for the frame shown below.

ANALYTICAL SOLUTION:

= 1.406

EI / L

= 55506.6 lb

Figure B7-1

B7A, B7B: Instability of Frames

See

List

GIVEN:

= 30 x 10

psi

= 1 in

= 25 in

= 1/12 in

COMPARISON OF RESULTS:

Theory

COSMOSM

SHELL4

SHELL6

(Curved)

SHELL6

(Assembled)

55506.6 lb 56280.8 lb 55364.9 lb

55732.1lb

Finite Element Model

Problem Sketch

Chapter 4 Buckling Analysis

4-10

COSMOSM Basic FEA System

TYPE:

Buckling analysis, axisymmetric shell element (SHELLAX).

REFERENCE:

Brush, D. O., and Almroth, B. O., “Buckling of Bars, Plates, and Shells,” McGraw-
Hill, Inc., New York, 1975, p. 164.

PROBLEM:

Find the buckling
load and deflection
mode for a cylindrical
shell that is simply
supported at its ends
and subjected to uni-
form lateral pressure.

GIVEN:

= 10 x 10

psi

= 0.2 in

= 20 in

= 0.3

ANALYTICAL SOLUTION:

B8: Instability of a Cylinder

See

List

COMPARISON OF RESULTS:

Theory

COSMOSM

106 psi

113.97 psi

Finite Element Model

Problem Sketch

Figure B8-1

COSMOSM Basic FEA System

4-11

Part 2 Verification Problems

TYPE:

Buckling analysis, shell (SHELL4) and beam (BEAM3D) elements.

REFERENCE:

Timoshenko, S. P., and Gere, J. M., “Theory of Elastic Stability,” 2nd edition,
McGraw-Hill Book Co., Inc., New York, p. 394, Table 9-16.

PROBLEM:

A simply supported rectangular plate is stiffened by a beam of rectangular cross-
section as shown in the figure. The stiffened plate is subjected to inplane pressure at
edges x = 0 and x = a. Determine the buckling pressure load.

COMPARISON OF RESULTS:

Figure B9-1

B9: Simply Supported Stiffened Plate

See

List

GIVEN:

ANALYTICAL SOLUTION:

Where:

β = a/b γ = EI

/ bD D = E(h

)

/ 12(1-

)

= b

δ = A

/ bh

= 30,000 kip/in

= 0

= 1 in

= 45.5 in

b = 42 in

= 0.42 in

= 10 in

Theory

COSMOSM

Difference

223.80 kip/in

232.53 kip/in

3.9%

Problem Sketch and Finite Element Model

Chapter 4 Buckling Analysis

4-12

COSMOSM Basic FEA System

TYPE:

Buckling analysis, beam elements (BEAM2D).

REFERENCE:

Timoshenko, S. P. and Gere J. M., “Theory of Elastic Stability,” McGraw-Hill Book
Co., New York, 1961.

Figure B10-1

B10: Stability of a Rectangular Frame

See

List

GIVEN:

= b = l00 in

= 1 in

= 1 in (beam cross section

height)

= 0.0833 in

= 1 x 10

psi

= 100 lb

COMPARISON OF RESULTS:

Theory

COSMOSM

1372.45 lb

1371.95 lb

ANALYTICAL SOLUTION:

= 16.47EI/L

= 1372.4451 lb

Problem Sketch

Finite Element Model

b/2

L/2

COSMOSM Basic FEA System

4-13

Part 2 Verification Problems

TYPE:

Buckling analysis, beam element (BEAM2D).

RFERENCE:

Roark, R. J. and Young, Y. C., “Formulas for Stress and Strain,” McGraw-Hill, New
York, l975, pp. 534.

PROBLEM:

Find the critical load and mode shape for the stepped column shown below.

Figure B11-1

B11: Buckling of a Stepped Column

See

List

GIVEN:

= 1000 mm

= 10,954 mm

= 15,492 mm

= E

= 68,950 MPa

= 0.3

= 1 x 10

= 2 x 10

= 0.5

COMPARISON OF RESULTS:

Theory

COSMOSM

554.6 KN

554.51 KN

ANALYTICAL SOLUTION:

= 0.326

/ (2L)

= 554,600 N

= 554.6 KN

Finite Element Model

Problem Sketch

E , A , I

Chapter 4 Buckling Analysis

4-14

COSMOSM Basic FEA System

TYPE:

Buckling analysis, composite shell element (SHELL4L).

REFERENCE:

Jones, “Mechanics of Composite Material,” McGraw-Hill Book Co., New York, p.
269.

PROBLEM:

Find the buckling load for [45,-45,45,-45] antisymmetric angle-ply laminated plate
under uniform axial compression p.

Figure B12-1

B12: Buckling Analysis of a Simply

Supported Composite Plate

See

List

GIVEN:

= b = 40 in

Σh

= 1 in

= 400,000 psi

= 10,000 psi

= 0.25

= G

=5,000 psi

= 1 lb/in

COMPARISON OF RESULTS:

Theory

COSMOSM

334.0 lb/in

345.36 lb/in

ANALYTICAL SOLUTION:

Approximate solution is given by graph
5-16 in the reference.

Problem Sketch and Finite Element Model

COSMOSM Basic FEA System

4-15

Part 2 Verification Problems

TYPE:

Buckling analysis, beam element (BEAM2D).

REFERENCE:

Timoshenko, S. P., and Gere, J. M., “Theory of Elastic Stability,” McGraw-Hill
Book Co., New York, 1961, pp. 125-128.

GIVEN:

= 1 in

= 4 in

b/b

= (x/a)

= 1 in

= 4 in

= 0.25

= 100 in

= 1 x 10

psi

= 1 in

ANALYTICAL SOLUTION:

= 1.678EI

/ L

= 6712 lb

COMPARISON OF RESULTS:

The Critical Load:

B13: Buckling of a Tapered Column

See

List

Theory

COSMOSM

6712 lb

6718 lb

Finite Element Model

Problem Sketch

SIDE VIEW

PLAN

b x

Figure B13-1

Chapter 4 Buckling Analysis

4-16

COSMOSM Basic FEA System

TYPE:

Linear buckling analysis using the nonaxisymmetric buckling mode option
(SHELLAX).

REFERENCE:

Sobel, L. H., “Effect of Boundary Conditions on the Stability of Cylinders Subject
to Lateral and Axial Pressures,” AIAA Journal, Vol. 2, No. 8, August, 1964, pp
1437-1440.

PROBLEM:

Find the buckling pressure for
the shown axisymmetric
clamped-clamped shell.

GIVEN:

= 1 in

= 0.3

= 4 in

= 10

psi

= 0.01 in

MODELING HINTS:

The cylindrical shell is
modeled with 20 uniform
elements. The starting harmonic number for which the buckling load is calculated is
set to 2. The minimum buckling load occurs at harmonic number 5 which
corresponds to mode shape 4 since the program started from harmonic 2.

B14: Buckling of Clamped Cylindrical Shell

Under External Pressure Using the

Nonaxisymmetric Buckling Mode Option

See

List

Finite Element Model

Problem Sketch

Figure B14-1

COSMOSM Basic FEA System

4-17

Part 2 Verification Problems

COMPARISON OF RESULTS:

Theory

COSMOSM

Harmonic Number

Critical Load

33.5 psi

35.0 psi

4-18

COSMOSM Basic FEA System

TYPE:

Linear buckling analysis using the nonaxisymmetric buckling mode option
(SHELLAX).

REFERENCE:

Timoshenko, S. P., and Gere, J. M., “Theory of Elastic Stability,” McGraw-Hill
Book Co., 1961.

PROBLEM:

Find the buckling load for
the simply-supported
cylindrical shell shown in
the figure below.

GIVEN:

= 10 in

= 16 in

ν (NUXY) = 0.3
E

= 10

psi

= 0.1 in

MODELING HINTS:

The cylindrical shell was modeled with 60 uniform elements. The starting harmonic
number for which the buckling load is calculated was set to 1. The solution stopped
at harmonic number 2 at which the minimum buckling load occurs. The number of
maximum iterations for the eigenvalue calculations was set to 100.

B15A, B15B: Buckling of Simply-Supported

Cylindrical Shell Under Axial Load

See

List

Finite Element Model